tag:blogger.com,1999:blog-47873945092723110492024-03-19T05:16:50.529-07:00Teoremas de geometríaDr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.comBlogger64125tag:blogger.com,1999:blog-4787394509272311049.post-20834140066212068402015-11-26T15:35:00.000-08:002015-11-26T15:38:27.669-08:00Teoremas de geometría<div class="separator" style="clear: both; text-align: center;">
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<br />Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com0tag:blogger.com,1999:blog-4787394509272311049.post-16926894852641618662014-01-04T15:10:00.001-08:002014-01-12T07:45:19.690-08:00Las medianas de un paralelogramo inscrito en una elipse son sus ejes conjugados
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<h2>
medi paralelog es eje conjuga</h2>
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<span style="font-size: x-small;">Néstor Martín Gulias, Creado con <a href="http://www.geogebra.org/" target="_blank">GeoGebra</a></span><br />
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Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com0tag:blogger.com,1999:blog-4787394509272311049.post-39206760368129006072012-06-09T01:54:00.001-07:002012-06-09T02:01:10.492-07:00Teorema de la ceviana<b><br />
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Si construimos una circunferencia de centro A y un punto exterior C desde el que trazamos las tangentes y a continuación hacemos una recta tangente a la circunferencia en un punto dado B, si unimos el punto de tangencia con el punto exterior C tenemos una recta CB que se llama ceviana.<br />
Si construimos la circunferencia de centro D inscrita a las tres tangentes tenemos que la ceviana corta a esta circunferencia en el punto H diametralmente opuesto al punto de tangencia F. Esto es debido a que las dos circunferencia son homotéticas y la homotecia conserva los ángulos. Si tomamos el punto medio I de los dos puntos de tangencia FB y trazamos una recta paralela a la ceviana tenemos que esta recta DI pasa por el centro D de la circunferencia inscrita a las tres tangentes.<br />
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<h2>Teorema de la ceviana</h2><br />
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name="useBrowserForJS" value="true" /><param name="allowRescaling" value="true" />Este es un Applet de Java creado con GeoGebra desde www.geogebra.org – Java no parece estar instalado Java en el equipo. Se aconseja dirigirse a www.java.com<br />
</applet></td></tr>
</tbody></table>Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com0tag:blogger.com,1999:blog-4787394509272311049.post-3411700577229105892012-06-09T01:51:00.003-07:002012-07-06T01:16:58.023-07:00T. de los triángulos semejantes en una inversión<b>Teorema de los triángulos semejantes en una inversión</b><br />
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</b><br />
Si construimos una circunferencia y un punto exterior C desde el que trazamos las tangentes, éstas interceptan a la circunferencia en dos puntos de tangencia DE que los unimos mediante una cuerda. Esta cuerda corta a la línea que une el centro de la circunferencia A y el punto exterior C en un punto I, este punto es el inverso del exterior C. <br />
Al construir por el centro de la circunferencia una recta paralela a la cuerda DE tenemos en la intersección con la circunferencia el punto F. Al unir F con el exterior C obtenemos un segmento que corta a la circunferencia en el punto H del que construimos su simétrico H’ respecto a la línea AC. <br />
Al unir el simétrico con F tenemos que corta a la línea AC en I. El triángulo AIF es semejante del triángulo AFC. Por tanto:<br />
AF/AI=AC/AF, siendo el segmento AF el radio de la circunferencia tenemos que AI.AC=AF.AF, AI.AC es igual al radio de la circunferencia al cuadrado.<br />
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<table border="0" style="width: 600px;"><tbody>
<tr><td><br />
<h2>T. de los triángulos semejantes en una inversión</h2><br />
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name="useBrowserForJS" value="true" /><param name="allowRescaling" value="true" />Este es un Applet de Java creado con GeoGebra desde www.geogebra.org – Java no parece estar instalado Java en el equipo. Se aconseja dirigirse a www.java.com </applet></td></tr>
</tbody></table>Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com0tag:blogger.com,1999:blog-4787394509272311049.post-25186791208956842612012-04-06T06:31:00.007-07:002012-05-06T00:58:13.279-07:00Teorema de la circunferencia focal<div class="separator" style="clear: both; text-align: center;"><a href="http://sistema-diedrico.blogspot.com.es/">http://sistema-diedrico.blogspot.com.es/</a> </div><div class="separator" style="clear: both; text-align: center;"><a href="http://curvas-conicas.blogspot.com.es/">http://curvas-conicas.blogspot.com.es/</a> </div><div class="separator" style="clear: both; text-align: center;"><br />
</div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjREv70yTia0qrkWAz_pvMuqrzPzV1xXkXBgmkbKMQo-BCV8KAjaijRsIZXoHO7MkCxbJwyHXKW2NFuiq5DCIUlFK2wq9HckSWg1jcw_wkUnuegUR6UidgPwN5sxHsJnJltxVeMiIhyOPk/s1600/Teoremas+de+la+c.+focal,.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="530" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjREv70yTia0qrkWAz_pvMuqrzPzV1xXkXBgmkbKMQo-BCV8KAjaijRsIZXoHO7MkCxbJwyHXKW2NFuiq5DCIUlFK2wq9HckSWg1jcw_wkUnuegUR6UidgPwN5sxHsJnJltxVeMiIhyOPk/s640/Teoremas+de+la+c.+focal,.jpg" width="640" /></a></div>La c. focal (centro en F2 y radio EH) de la elipse inscrita en un polígono inscrito, contiene a los simétricos de su otro foco F1 tomando los lados del polígono como ejes de simetría.<br />
Los puntos de tangencia del polígono y de la elipse unidos a los vértices NKO definen segmentos (en verde) que se cortan en un punto P.<br />
Al moverse K sobre la circunferencia focal, manteniéndose siempre los lados del polígono NOK tangentes a la elipse y N O también incidentes sobre la circunferencia, P se mueve describiendo otra elipse proporcional (de igual excentricidad) cuyo eje mayor coincide con el de la elipse EFGH.<br />
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<title>Teoremas de la c. focal - GeoGebra Hoja Dinámica</title><br />
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<h2>Teoremas de la c. focal </h2>Mover los puntos K F1 F2 para observar los cambios de la elipse que describe P.<br />
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<title>Teoremas de la c. focal-rastro - GeoGebra Hoja Dinámica</title><br />
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<h2>Teoremas de la c. focal-rastro</h2>Mover el punto K para ver la formación de la nueva elipse y los focos F1 F2 para ver que la elipse es proporcional a la original.<br />
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/><param name="image" value="http://www.geogebra.org/webstart/loading.gif" /><param name="boxborder" value="false" /><param name="centerimage" value="true" /><param name="java_arguments" value="-Xmx512m -Djnlp.packEnabled=true" /><param name="cache_archive" value="geogebra.jar, geogebra_main.jar, geogebra_gui.jar, geogebra_cas.jar, geogebra_export.jar, geogebra_properties.jar" /><param name="cache_version" value="3.2.47.0, 3.2.47.0, 3.2.47.0, 3.2.47.0, 3.2.47.0, 3.2.47.0" /><param name="framePossible" value="false" /><param name="showResetIcon" value="false" /><param name="showAnimationButton" value="true" /><param name="enableRightClick" value="false" /><param name="errorDialogsActive" value="true" /><param name="enableLabelDrags" value="false" /><param name="showMenuBar" value="false" /><param name="showToolBar" value="false" /><param name="showToolBarHelp" value="false" /><param name="showAlgebraInput" value="false" /><param name="allowRescaling" value="true" />This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com<br />
</applet><br />
<br />
</td></tr>
</tbody></table>Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com0tag:blogger.com,1999:blog-4787394509272311049.post-79647243690752460262012-04-02T13:58:00.000-07:002012-04-02T13:58:24.773-07:00Teorema de las simetrías del ortocentro en un triángulo inscrito.<title>Simétricos del ortocentro - GeoGebra Hoja Dinámica</title><br />
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Los lados de un triángulo inscrito en una circunferencia son ejes de simetría entre el ortocentro y sus homólogos sobre la circunferencia.<br />
<table border="0" style="width: 600px;"><tbody>
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<h2>Simétricos del ortocentro</h2><br />
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<applet archive="geogebra.jar" code="geogebra.GeoGebraApplet" codebase="http://www.geogebra.org/webstart/3.2/unsigned/" height="469" mayscript="true" name="ggbApplet" title="Java" width="499"><br />
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</applet></td></tr>
</tbody></table>Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com0tag:blogger.com,1999:blog-4787394509272311049.post-19996464023843124372012-04-02T12:50:00.000-07:002012-04-02T12:50:37.911-07:00Teorema del primer y segundo punto de Fermat<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgi7DdCQ1qgNlqcPXQWsQ2xMM3uk-W6l4qymZHdNsaAuoUlozeLByGA_AFgXZQVH3JWBuIC01JS_euveTy7tRU3YjQ3V317j-NGfrcSzfuE5NgOp3_Cj5RkmQu3H08cBNpcQP4wnrMi6dA/s1600/1%C2%BA+punto+de+Fermat.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="466" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgi7DdCQ1qgNlqcPXQWsQ2xMM3uk-W6l4qymZHdNsaAuoUlozeLByGA_AFgXZQVH3JWBuIC01JS_euveTy7tRU3YjQ3V317j-NGfrcSzfuE5NgOp3_Cj5RkmQu3H08cBNpcQP4wnrMi6dA/s640/1%C2%BA+punto+de+Fermat.jpg" width="640" /></a></div><br />
<title>1º punto de Fermat. - GeoGebra Hoja Dinámica</title><br />
Al unir los puntos de tangencia JKL de 3 circunferencias exinscritas a un triángulo ABC, con los vértices opuestos CAB respectivamente, tenemos 3 líneas que se cortan en punto M.<br />
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<h2>1º punto de Fermat.</h2><br />
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<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjXpjXfq6aHSC5KciVhKb1T2mT4eYX0muhyWLMkZD-UmvTBTZflUXS4fgkwB9qbtDzLRwyiJzF__DzUWc5V4KIO-yEE3XTTNH9FXNq6v0xO3U37w0yF9sFi62m9xEjXXF66pBy-fKhC3GM/s1600/2%C2%BA+punto+de+Fermat..jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="466" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjXpjXfq6aHSC5KciVhKb1T2mT4eYX0muhyWLMkZD-UmvTBTZflUXS4fgkwB9qbtDzLRwyiJzF__DzUWc5V4KIO-yEE3XTTNH9FXNq6v0xO3U37w0yF9sFi62m9xEjXXF66pBy-fKhC3GM/s640/2%C2%BA+punto+de+Fermat..jpg" width="640" /></a></div><br />
<title>2º punto de Fermat - GeoGebra Hoja Dinámica</title><br />
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Si sobre el triángulo anterior ABC que definen las 3 líneas hacemos triángulos equiláteros, al unir sus vértices<br />
P con A, O con B y N con C, tenemos tres rectas incidentes en un vértice Q.<br />
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<h2>2º punto de Fermat</h2><br />
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/> <param name="image" value="http://www.geogebra.org/webstart/loading.gif" /> <param name="boxborder" value="false" /> <param name="centerimage" value="true" /> <param name="java_arguments" value="-Xmx512m -Djnlp.packEnabled=true" /> <param name="cache_archive" value="geogebra.jar, geogebra_main.jar, geogebra_gui.jar, geogebra_cas.jar, geogebra_export.jar, geogebra_properties.jar" /> <param name="cache_version" value="3.2.47.0, 3.2.47.0, 3.2.47.0, 3.2.47.0, 3.2.47.0, 3.2.47.0" /> <param name="framePossible" value="false" /> <param name="showResetIcon" value="false" /> <param name="showAnimationButton" value="true" /> <param name="enableRightClick" value="false" /> <param name="errorDialogsActive" value="true" /> <param name="enableLabelDrags" value="false" /> <param name="showMenuBar" value="false" /> <param name="showToolBar" value="false" /> <param name="showToolBarHelp" value="false" /> <param name="showAlgebraInput" value="false" /> <param name="allowRescaling" value="true" />This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com<br />
</applet><br />
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</td></tr>
</tbody></table></td></tr>
</tbody></table>Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com0tag:blogger.com,1999:blog-4787394509272311049.post-56903960654762026622012-04-02T12:39:00.000-07:002012-04-02T12:39:53.026-07:00Teorema del punto de Vecten<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj0dcdWdTaUDw6S5FJteFzY8saOR-bNsPaBo9pq4ghj07H-Fag-fUIxYBhkKb5ivsjPYbz1JXRIwNuXMO8wPBfEIIcFFI8BWBR5Lw_MFC1ZOHKJTOH01yi4moezuvRGJMXrA3fYognXyck/s1600/punto+de+Vecten.jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="396" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj0dcdWdTaUDw6S5FJteFzY8saOR-bNsPaBo9pq4ghj07H-Fag-fUIxYBhkKb5ivsjPYbz1JXRIwNuXMO8wPBfEIIcFFI8BWBR5Lw_MFC1ZOHKJTOH01yi4moezuvRGJMXrA3fYognXyck/s640/punto+de+Vecten.jpg" width="640" /></a></div>Si sobre los lados de un triángulo ABC construimos cuadrados de los que tomamos sus centros JKL y los alineamos con el vértice opuesto del lado del triángulo donde se apoyan (CAB respectivamente), las 3 líneas que se forman JC KA BL se cortan en un punto M.<br />
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<title>punto de Vecten - GeoGebra Hoja Dinámica</title><br />
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<h2>Teorema del punto de Vecten</h2><br />
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</applet><br />
<br />
<br />
<br />
</td></tr>
</tbody></table>Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com0tag:blogger.com,1999:blog-4787394509272311049.post-10804783128559378562012-04-01T15:23:00.000-07:002012-04-01T15:23:02.697-07:00T. del triángulo equilátero inscrito<title>T. del triángulo equilátero inscrito - - GeoGebra Hoja Dinámica</title><br />
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Si un triángulo ABC está inscrito en una circunferencia sobre la que tomamos un punto E desde el que trazamos segmentos hasta los vértices, la longitud de un segmento AE (azul) es la misma que la longitud de los otros dos segmentos (en rosa) sumados CE+EB.<br />
AE=CE+EB.<br />
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<table border="0" style="width: 777px;"><tbody>
<tr><td><br />
<h2>T. del triángulo equilátero inscrito -</h2><br />
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/> <param name="image" value="http://www.geogebra.org/webstart/loading.gif" /> <param name="boxborder" value="false" /> <param name="centerimage" value="true" /> <param name="java_arguments" value="-Xmx512m -Djnlp.packEnabled=true" /> <param name="cache_archive" value="geogebra.jar, geogebra_main.jar, geogebra_gui.jar, geogebra_cas.jar, geogebra_export.jar, geogebra_properties.jar" /> <param name="cache_version" value="3.2.47.0, 3.2.47.0, 3.2.47.0, 3.2.47.0, 3.2.47.0, 3.2.47.0" /> <param name="framePossible" value="false" /> <param name="showResetIcon" value="false" /> <param name="showAnimationButton" value="true" /> <param name="enableRightClick" value="false" /> <param name="errorDialogsActive" value="true" /> <param name="enableLabelDrags" value="false" /> <param name="showMenuBar" value="false" /> <param name="showToolBar" value="false" /> <param name="showToolBarHelp" value="false" /> <param name="showAlgebraInput" value="false" /> <param name="allowRescaling" value="true" />This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com<br />
</applet></td></tr>
</tbody></table>Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com0tag:blogger.com,1999:blog-4787394509272311049.post-77931200993719167862012-03-30T16:53:00.003-07:002012-04-06T12:54:12.667-07:00Teorema de Soddy<div style="text-align: -webkit-left;">Existen únicamente 2 circunferencias tangentes a 3 circunferencias tangentes entre sí, se les llama circunferencias de Soddy.</div><span style="color: #4d469c; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif;"><span style="line-height: 18px;"><br />
</span></span><br />
<a href="http://tangencias-y-enlaces.blogspot.com/" style="background-color: #e3a327; color: #4d469c; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: 13px; line-height: 18px; text-decoration: none;">http://tangencias-y-enlaces.blogspot.com/</a> <br />
<a href="http://tangencias-inversion.blogspot.com/" style="background-color: #e3a327; color: #4d469c; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: 13px; line-height: 18px; text-decoration: none;">http://tangencias-inversion.blogspot.com/</a> <br />
<a href="http://tangencias-potencia.blogspot.com/" style="background-color: #e3a327; color: #4d469c; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: 13px; line-height: 18px; text-decoration: none;">http://tangencias-potencia.blogspot.com/</a> <br />
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<table border="0" style="width: 910px;"><tbody>
<tr><td><h2>Teorema de Soddy</h2><br />
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<applet archive="geogebra.jar" code="geogebra.GeoGebraApplet" codebase="http://www.geogebra.org/webstart/3.2/unsigned/" height="555" mayscript="true" name="ggbApplet" title="Java" width="910"><br />
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/><param name="image" value="http://www.geogebra.org/webstart/loading.gif" /><param name="boxborder" value="false" /><param name="centerimage" value="true" /><param name="java_arguments" value="-Xmx512m -Djnlp.packEnabled=true" /><param name="cache_archive" value="geogebra.jar, geogebra_main.jar, geogebra_gui.jar, geogebra_cas.jar, geogebra_export.jar, geogebra_properties.jar" /><param name="cache_version" value="3.2.47.0, 3.2.47.0, 3.2.47.0, 3.2.47.0, 3.2.47.0, 3.2.47.0" /><param name="framePossible" value="false" /><param name="showResetIcon" value="false" /><param name="showAnimationButton" value="true" /><param name="enableRightClick" value="false" /><param name="errorDialogsActive" value="true" /><param name="enableLabelDrags" value="false" /><param name="showMenuBar" value="false" /><param name="showToolBar" value="false" /><param name="showToolBarHelp" value="false" /><param name="showAlgebraInput" value="false" /><param name="allowRescaling" value="true" />This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com<br />
</applet></td></tr>
</tbody></table>Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com0tag:blogger.com,1999:blog-4787394509272311049.post-59765665049455844942012-03-30T16:22:00.006-07:002012-04-11T03:30:28.273-07:00Teoremas de Mikami y Kobayashi<title>Y. Mikami y T. Kobayashi - GeoGebra Hoja Dinámica</title><br />
Al construir un cuadrilátero BCDE inscrito en una circunferencia (en rosa) y sus diagonales (en verde), determinamos los triángulos BCE, EDB, ECD y DBC en los que hacemos 4 circunferencias inscritas, los centros de éstas definen siempre un rectángulo.<br />
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<h2>Teorema de Y. Mikami y T. Kobayashi</h2><br />
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/><param name="image" value="http://www.geogebra.org/webstart/loading.gif" /><param name="boxborder" value="false" /><param name="centerimage" value="true" /><param name="java_arguments" value="-Xmx512m -Djnlp.packEnabled=true" /><param name="cache_archive" value="geogebra.jar, geogebra_main.jar, geogebra_gui.jar, geogebra_cas.jar, geogebra_export.jar, geogebra_properties.jar" /><param name="cache_version" value="3.2.47.0, 3.2.47.0, 3.2.47.0, 3.2.47.0, 3.2.47.0, 3.2.47.0" /><param name="framePossible" value="false" /><param name="showResetIcon" value="false" /><param name="showAnimationButton" value="true" /><param name="enableRightClick" value="false" /><param name="errorDialogsActive" value="true" /><param name="enableLabelDrags" value="false" /><param name="showMenuBar" value="false" /><param name="showToolBar" value="false" /><param name="showToolBarHelp" value="false" /><param name="showAlgebraInput" value="false" /><param name="allowRescaling" value="true" />This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com<br />
</applet></td></tr>
</tbody></table><br />
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Al triangular el cuadrilátero de distintas formas, se tiene que la suma de los radios de las circunferencias inscritas en los triángulos es siempre el mismo.<br />
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En el dibujo se han hecho 2 triangulaciones distintas, las de los círculos rojos y las de los verdes, la otra es común a ambas. Observamos por un grupo de traslaciones que la suma de los radios de las verdes es igual que la suma de los radios de las rojas.</div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjfKjD-dSmiF6KjIVEfMVfRWHM3lU_yGlqel5iycdx7sL3gCibN_zAbhYc5pyORZo5p8Da47nLwaGzfdrOzhCM8ia35BHuQp1xD6Of8ew3tTh-Btn1HSVkUUca74aYPiZMgQTrkPw-CyeM/s1600/Y.+Mikami+y+T.+Kobayashi+2..jpg" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="464" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjfKjD-dSmiF6KjIVEfMVfRWHM3lU_yGlqel5iycdx7sL3gCibN_zAbhYc5pyORZo5p8Da47nLwaGzfdrOzhCM8ia35BHuQp1xD6Of8ew3tTh-Btn1HSVkUUca74aYPiZMgQTrkPw-CyeM/s640/Y.+Mikami+y+T.+Kobayashi+2..jpg" width="640" /></a></div><div><br />
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<table border="0" style="width: 642px;"><tbody>
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<h2>T. de Y. Mikami y T. Kobayashi 2</h2>Al mover los puntos siempre queda un rectángulo por lo que queda demostrada la igualdad.<br />
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/><param name="image" value="http://www.geogebra.org/webstart/loading.gif" /><param name="boxborder" value="false" /><param name="centerimage" value="true" /><param name="java_arguments" value="-Xmx512m -Djnlp.packEnabled=true" /><param name="cache_archive" value="geogebra.jar, geogebra_main.jar, geogebra_gui.jar, geogebra_cas.jar, geogebra_export.jar, geogebra_properties.jar" /><param name="cache_version" value="3.2.47.0, 3.2.47.0, 3.2.47.0, 3.2.47.0, 3.2.47.0, 3.2.47.0" /><param name="framePossible" value="false" /><param name="showResetIcon" value="false" /><param name="showAnimationButton" value="true" /><param name="enableRightClick" value="false" /><param name="errorDialogsActive" value="true" /><param name="enableLabelDrags" value="false" /><param name="showMenuBar" value="false" /><param name="showToolBar" value="false" /><param name="showToolBarHelp" value="false" /><param name="showAlgebraInput" value="false" /><param name="allowRescaling" value="true" />This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com<br />
</applet><br />
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</td></tr>
</tbody></table>Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com0tag:blogger.com,1999:blog-4787394509272311049.post-9948885496451508042012-03-22T06:06:00.005-07:002012-06-23T00:59:39.427-07:00Teorema de los triángulos podales equivalentes<div class="separator" style="clear: both; text-align: left;">Si dibujamos un triángulo cualquiera abc inscrito en una circunferencia g y un punto N exterior o interior a la misma desde el que trazamos rectas perpendiculares a los lados del triángulo obtenemos tres puntos IJH que unidos determinan un triángulo llamado podal.</div><div class="separator" style="clear: both; text-align: left;">Teorema: si construimos una circunferencia cualquiera t concéntrica a la anterior g que pase por N y a continuación trasladamos este punto a lo largo de toda la circunferencia, todos los triángulos definidos por la intersección de las perpendiculares a los lados del triángulo son equivalentes. Por ejemplo, el punto N al trasladarse a su nueva posición M define el nuevo triángulo AFG que tiene la misma área que el anterior IJH.</div><div class="separator" style="clear: both; text-align: left;"><br />
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</div><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjk4AADXorBzLu0-af45K2U_e-GNRXUKJWKT_Gmm0Dbex0FX50bWZgSMBrj56BKMvuxfGpJYwKbVmwPiVqMoxcf7Aj5O67ukmbfEcIno6L8OTLOUkLRK0irAilWirdDIbfxVs1B7jjFRNg/s1600/tri%C3%A1ngulos+podales+equivalentes.bmp" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="376" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjk4AADXorBzLu0-af45K2U_e-GNRXUKJWKT_Gmm0Dbex0FX50bWZgSMBrj56BKMvuxfGpJYwKbVmwPiVqMoxcf7Aj5O67ukmbfEcIno6L8OTLOUkLRK0irAilWirdDIbfxVs1B7jjFRNg/s640/tri%C3%A1ngulos+podales+equivalentes.bmp" width="640" /></a></div><br />
En la figura siguiente observamos un triángulo DEF inscrito en una circunferencia de radio 3 y un triángulo podal verde IJH cuyos vértices son las intersecciones de las perpendiculares por G (un punto cualquiera) con los lados de DEF o de sus prolongaciones. Si hacemos una circunferencia concéntrica a la anterior que pase por G, al trasladar G a lo largo de toda la circunferencia permaneciendo ésta invariable en su diámetro, genera nuevos triángulos que tienen todos la misma área.<br />
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<h2>T. de los triángulos podales</h2><br />
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name="useBrowserForJS" value="true" /><param name="allowRescaling" value="true" />Este es un Applet de Java creado con GeoGebra desde www.geogebra.org – Java no parece estar instalado Java en el equipo. Se aconseja dirigirse a www.java.com<br />
</applet></td></tr>
</tbody></table>Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com0tag:blogger.com,1999:blog-4787394509272311049.post-49594871997365217822012-03-20T12:44:00.002-07:002012-03-20T12:48:45.101-07:00Teorema de Von Aubel<title>Teorema de los 4 cuadrados - GeoGebra Hoja Dinámica</title><br />
<br />
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<style type="text/css">
<!--body { font-family:Arial,Helvetica,sans-serif; margin-left:40px }-->
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Al construir cuadrados sobre los lados de un cuadrilátero y tomar sus puntos centrales opuestos y unirlos mediante segmentos obtenemos siempre dos rectas perpendiculares.<br />
<table border="0" style="width: 600px;"><tbody>
<tr><td><br />
<h2>Teorema de Von Aubel</h2><br />
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</applet></td></tr>
</tbody></table>Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com0tag:blogger.com,1999:blog-4787394509272311049.post-81891054880123356082012-03-20T12:04:00.002-07:002012-03-22T09:32:48.493-07:00Teorema de la bisectriz interior de un triángulo<title>Teorema de la bisectriz interior de un triángulo - GeoGebra Hoja Dinámica</title><br />
El segmento correspondiente a la bisectriz interior de un triángulo elevado al cuadrado es igual al producto de dos lados menos el producto de los dos segmentos de la base del triángulo que divide la bisectriz.<br />
Si restamos al área del rectángulo mayor el área del rectángulo menor obtenemos el área del cuadrado.<br />
<table border="0" style="width: 600px;"><tbody>
<tr><td><br />
<h2>Teorema de la bisectriz interior de un triángulo</h2><br />
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/><param name="image" value="http://www.geogebra.org/webstart/loading.gif" /><param name="boxborder" value="false" /><param name="centerimage" value="true" /><param name="java_arguments" value="-Xmx512m -Djnlp.packEnabled=true" /><param name="cache_archive" value="geogebra.jar, geogebra_main.jar, geogebra_gui.jar, geogebra_cas.jar, geogebra_export.jar, geogebra_properties.jar" /><param name="cache_version" value="3.2.47.0, 3.2.47.0, 3.2.47.0, 3.2.47.0, 3.2.47.0, 3.2.47.0" /><param name="framePossible" value="false" /><param name="showResetIcon" value="false" /><param name="showAnimationButton" value="true" /><param name="enableRightClick" value="false" /><param name="errorDialogsActive" value="true" /><param name="enableLabelDrags" value="false" /><param name="showMenuBar" value="false" /><param name="showToolBar" value="false" /><param name="showToolBarHelp" value="false" /><param name="showAlgebraInput" value="false" /><param name="allowRescaling" value="true" />This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com<br />
</applet></td></tr>
</tbody></table><br />
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjoL1RoNuhnFoRpx_kHrkQz0Gf0TdIsjL2GHTSHyDQEmj-UP3es80rr5alSlnPG9zGv7B8tWCI1igqZeGoqvtK2_WZlkCorDaqV3jqNb6EMl5Q1tvRLdureLv7lAxeVK74jH0L1P2RKCa0/s1600/Teorema+de+la+bisectriz+interior+de+un+tri%C3%A1ngulo.bmp" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="376" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjoL1RoNuhnFoRpx_kHrkQz0Gf0TdIsjL2GHTSHyDQEmj-UP3es80rr5alSlnPG9zGv7B8tWCI1igqZeGoqvtK2_WZlkCorDaqV3jqNb6EMl5Q1tvRLdureLv7lAxeVK74jH0L1P2RKCa0/s640/Teorema+de+la+bisectriz+interior+de+un+tri%C3%A1ngulo.bmp" width="640" /></a></div><div class="separator" style="clear: both; text-align: left;">En la figura, el área del rectángulo de color rosa menos el área del rectángulo naranja es igual al área del cuadrado amarillo.</div><div><br />
</div>Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com0tag:blogger.com,1999:blog-4787394509272311049.post-21676556546342431332012-03-20T11:11:00.002-07:002012-03-20T11:30:30.630-07:00Teorema de De Gua<div class="separator" style="clear: both; text-align: left;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgN_uY6HISQsoyg9t39PWeJ4nz_IFPyIPhgsu7wEXGKyuVqis_I29OTfteMxEm_lnaPnMgTEtddnhnnHMbnK-Ig45XP-VOy79kC0byyerHxT3guUtssawsty6_hJ8acHie5GCdR5ljr8uk/s1600/Teorema+de+De+Gua3.bmp" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="392" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgN_uY6HISQsoyg9t39PWeJ4nz_IFPyIPhgsu7wEXGKyuVqis_I29OTfteMxEm_lnaPnMgTEtddnhnnHMbnK-Ig45XP-VOy79kC0byyerHxT3guUtssawsty6_hJ8acHie5GCdR5ljr8uk/s640/Teorema+de+De+Gua3.bmp" width="640" /></a></div><div class="separator" style="clear: both; text-align: left;">El teorema expresa la relación que existe entre los cuatro volúmenes prismáticos de la figura. </div><div class="separator" style="clear: both; text-align: left;">La suma de los volúmenes de los tres prismas amarillos es igual al volumen del prisma rojo. Si tenemos una pirámide que es un triedro trirrectángulro rectángulo, que quiere decir que el vértice superior es la esquina de un cubo, o sea que todas las caras tienen en ese vértice 90° y calculamos el área de cada cara aplicando luego el cuadrado de la misma, de esta forma obtenemos los prismas que aparecen dibujados y desplazados sobre cada cara. Si a continuación calculamos el área de la base de la pirámide y la multiplicamos por sí misma (elevamos su valor al cuadrado) obtenemos un volumen de la misma capacidad que la suma de los otros tres prismas amarillos. </div><div class="separator" style="clear: both; text-align: left;">Para calcular el área al cuadrado de una figura, multiplicamos el número por sí mismo, esto en el espacio no es más que darle al prisma la misma altura que el valor del área.</div><div class="separator" style="clear: both; text-align: left;">Ejemplo: en las tres aristas de la pirámide que definen las tres caras ABC, miden 3,4 y 5 unidades. La intersección de los planos AB es de cuatro unidades, la de los planos BC es de tres unidades y la de los planos AC es de cinco unidades. </div><div class="separator" style="clear: both; text-align: left;">El área de la primera B es 3 × 4 igual a 12 partido entre dos tiene por valor seis. </div><div class="separator" style="clear: both; text-align: left;">El área de la cara C es 3 × 5 igual a 15/2 tenemos 7,5. </div><div class="separator" style="clear: both; text-align: left;">El área de la cara A es 5 × 4 = 20 dividido entre 2 es 10. </div><div class="separator" style="clear: both; text-align: left;">El área de la base es 13,86, resultado de coger un lado como base y multiplicarlo por la altura y partirlo por dos.</div><div class="separator" style="clear: both; text-align: left;">Esta última cara de área 13,86 multiplicada por sí misma (elevado al cuadrado) nos da un valor de 192,24, el prisma recto por tanto tiene una altura de 13,86. Éste valor es igual a las otras áreas al cuadrado sumadas: 6 al cuadrado mas 7,5 al cuadrado más 10 al cuadrado es 36 + 56, 25 + 100, respectivamente que es igual a 192 ,24. </div>Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com0tag:blogger.com,1999:blog-4787394509272311049.post-46346305389571376802012-03-19T12:57:00.000-07:002012-03-19T12:57:15.661-07:00Teorema de Lambert<title>Teorema de Lambert - GeoGebra Hoja Dinámica</title><br />
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</span></div>La circunferencia circunscrita a un triángulo determinado por tres tangentes a la parábola incide en el foco de la misma.<br />
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<h2>Teorema de Lambert</h2><br />
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</applet></td></tr>
</tbody></table>Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com0tag:blogger.com,1999:blog-4787394509272311049.post-69444339289028520112012-03-19T12:26:00.000-07:002012-03-19T12:26:25.655-07:00Teorema de Viviani<title>Teorema de Viviani - GeoGebra Hoja Dinámica</title><br />
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Si sumamos las distancias desde un punto a los lados de un triángulo equilátero tenemos que la longitud obtenida es igual a la altura del triángulo. El teorema es válido para todos los polígonos que tienen lados iguales y ángulos iguales.<br />
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<h2>Teorema de Viviani</h2><br />
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</applet></td></tr>
</tbody></table>Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com0tag:blogger.com,1999:blog-4787394509272311049.post-68168113636836638122012-03-19T11:46:00.004-07:002012-06-23T01:08:08.753-07:00Teorema de la bisectriz<title>Teorema de la bisectriz - GeoGebra Hoja Dinámica</title><br />
<span style="background-color: white; font-family: Georgia, serif; font-size: 13px; line-height: 19px;">En un triángulo, la razón entre dos lados (AB es a AC) es igual a la razón de las partes (BD es a CD) en las que queda dividido el tercer lado por la bisectriz AD de ángulo BAC (entre AB y AC).</span><br />
<br />
AB/AC= BD/CD <br />
<br />
AB.CD=AC.BD, por tanto los dos rectángulos rosa y azul son equivalentes, que quiere decir que tienen la misma área.<br />
<br />
<a href="http://figuras-equivalentes.blogspot.com.es/">http://figuras-equivalentes.blogspot.com.es/</a><br />
<br />
<table border="0" style="width: 600px;"><tbody>
<tr><td><br />
<h2>Teorema de la bisectriz</h2><br />
<br />
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created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com<br />
</applet></td></tr>
</tbody></table><br />
<br />
<br />
<br />
<title>Demostracción del teorema de la bisectriz - GeoGebra Hoja Dinámica</title><br />
<b><br />
</b><br />
<b><br />
</b><br />
<style type="text/css">
<!--body { font-family:Arial,Helvetica,sans-serif; margin-left:40px }-->
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<b>Demostración del teorema de la bisectriz</b><br />
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<div class="MsoNormal"><span style="line-height: 115%;">Según el teorema de la bisectriz, el segmento AB es al segmento AC como el segmento BD es al segmento DC. Pero si construimos las dos circunferencias del dibujo y colocamos los radios BD DC alineados con los segmentos AB AC, observamos que efectivamente el segmento BE es al segmento CF como el segmento AB es al segmento AC. Todo ello queda demostrado por ser los segmentos DC EF paralelos.<o:p></o:p></span></div><br />
<table border="0" style="width: 600px;"><tbody>
<tr><td><br />
<h2>Demostracción del teorema de la bisectriz</h2><br />
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</tbody></table>Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com0tag:blogger.com,1999:blog-4787394509272311049.post-67859835908050766442012-03-19T11:18:00.001-07:002012-03-20T00:30:06.122-07:00Teorema de Casey<span style="font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: medium;"><span style="line-height: 25px;"><br />
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</div><dl style="background-color: white; font-family: sans-serif; font-size: 13px; line-height: 19px; margin-bottom: 0.5em; margin-top: 0.2em;"><div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEil-UJ0f641UP_NlDgxIzIGR-GC8jzfcj11uqp7UieWewqtcmEAHSpnWMXbX24QWaSZXrTXcoty4daYaN0hPZCNStNSk1HBkwMXLTpEBGEb7hgnv4ye-Eysvc1TNaOyXYsMnc51l_bardE/s1600/Teorema+de+Casey.bmp" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="376" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEil-UJ0f641UP_NlDgxIzIGR-GC8jzfcj11uqp7UieWewqtcmEAHSpnWMXbX24QWaSZXrTXcoty4daYaN0hPZCNStNSk1HBkwMXLTpEBGEb7hgnv4ye-Eysvc1TNaOyXYsMnc51l_bardE/s640/Teorema+de+Casey.bmp" width="640" /></a></div><dd style="line-height: 1.5em; margin-bottom: 0.1em; margin-left: 1.6em; margin-right: 0px;">En cuatro circunferencias interiores tangentes a otra se tiene que el producto del las tangentes diagonales (segmentos en color verde) es igual a la suma de los productos de cada par de tangentes opuestas a cada par de circunferencias (segmentos en color rojo y segmentos en color azul).
En síntesis: el área del rectángulo verde es igual a la suma de las áreas de los rectángulos rosa y azul.</dd><dd style="line-height: 1.5em; margin-bottom: 0.1em; margin-left: 1.6em; margin-right: 0px;">Cuando las circunferencias se reducen a un punto tenemos el teorema de Ptolomeo.</dd></dl>Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com0tag:blogger.com,1999:blog-4787394509272311049.post-56054509612890416032012-03-19T10:22:00.001-07:002012-03-29T05:44:46.665-07:00Teorema de Steiner-LehmusTodo triángulo que tiene iguales 2 bisectrices es necesariamente isósceles.<br />
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<title>Teorema de Steiner-Lehmus - GeoGebra Hoja Dinámica</title><br />
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<h2>Teorema de Steiner-Lehmus</h2><br />
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</tbody></table>Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com0tag:blogger.com,1999:blog-4787394509272311049.post-75240123663962492562012-03-19T10:02:00.001-07:002012-03-19T13:53:22.297-07:00Teorema de Pick<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjgr5c6KKDvGjDGWWLOko5EvJWddLpgDhWTHqPbeJeuwt4N_9DuMtPW6lK3coOI9OEY4BDMhbITHq6sQCqfmXZvFWYErRfmpX1PFfTGwjscfpskPmWNDhmkryLDVk04x2EjgGlsssFl8Eg/s1600/teorema+de+Pick.bmp" imageanchor="1" style="margin-left: 1em; margin-right: 1em;"><img border="0" height="390" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEjgr5c6KKDvGjDGWWLOko5EvJWddLpgDhWTHqPbeJeuwt4N_9DuMtPW6lK3coOI9OEY4BDMhbITHq6sQCqfmXZvFWYErRfmpX1PFfTGwjscfpskPmWNDhmkryLDVk04x2EjgGlsssFl8Eg/s640/teorema+de+Pick.bmp" width="640" /></a></div><br />
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<tr><td><b>Para todo polígono de una pieza y sin agujeros, el número de puntos en el interior, más el número de puntos en el borde partido por 2, menos 1, es igual al área del polígono.</b><br />
Se le llama números enteros aquellos que tienen coordenadas enteras que quiere decir que están dentro de la superficie de la figura si son interiores y que pertenecen a los lados del polígono si están en el borde del polígono.</td></tr>
</tbody></table>Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com0tag:blogger.com,1999:blog-4787394509272311049.post-17696950811032061352012-03-18T15:49:00.004-07:002012-03-25T01:38:31.840-07:00Teorema de Pohlke<span style="color: grey; font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif;"><span style="font-size: 11px; line-height: 18px;"><br />
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<b>Dadas 3 líneas en el plano X'Y'Z' no coincidentes e incidentes en un punto, existe un triedro trirrectángulo XYZ en el espacio que puede transformarse en la tres líneas por proyección.</b><br />
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Para determinar el triedro que cumple esa condición se hace la pirámide de triedros trirrectángulos que se apoya sobre los tres ejes y su triángulo fundamental (intersección de la pirámide y el plano PC).<br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi5Ly8N6556xUmAgss4W6YMWHQ_POMQdnso-pWe0-8g4PH8HRLlE7mmbIbihLhmAGxsq_xg-7M7xSqp0mWoamOmoGRiVaZEBA8qDOI57Ah-UB5NKWG6Mr6uzcHA1asFZftPVBLu5xGXZ6Ti/s1600/3+axo.jpg" style="background-color: white; color: #26ff00; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: 13px; line-height: 18px;"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5599859007610311794" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEi5Ly8N6556xUmAgss4W6YMWHQ_POMQdnso-pWe0-8g4PH8HRLlE7mmbIbihLhmAGxsq_xg-7M7xSqp0mWoamOmoGRiVaZEBA8qDOI57Ah-UB5NKWG6Mr6uzcHA1asFZftPVBLu5xGXZ6Ti/s320/3+axo.jpg" style="-webkit-box-shadow: rgba(0, 0, 0, 0.0976563) 1px 1px 5px; background-attachment: initial; background-clip: initial; background-image: initial; background-origin: initial; border-bottom-color: rgb(238, 238, 238); border-bottom-style: solid; border-bottom-width: 1px; border-color: initial; border-image: initial; border-left-color: rgb(238, 238, 238); border-left-style: solid; border-left-width: 1px; border-right-color: rgb(238, 238, 238); border-right-style: solid; border-right-width: 1px; border-top-color: rgb(238, 238, 238); border-top-style: solid; border-top-width: 1px; border-width: initial; box-shadow: rgba(0, 0, 0, 0.0976563) 1px 1px 5px; float: left; height: 159px; margin-bottom: 10px; margin-left: 0px; margin-right: 10px; margin-top: 0px; padding-bottom: 5px; padding-left: 5px; padding-right: 5px; padding-top: 5px; position: relative; width: 320px;" /></a><br />
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<span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: 13px; line-height: 18px;">En la figura podemos observar el fundamento del sistema axonométrico, tres ejes cartesianos, X Y Z se proyectan sobre un plano mediante líneas paralelas y de forma ortogonal al mismo. Estos tres ejes son la esquina de un cubo (forman entre sí dos a dos 90°, lo que en geometría se llama un triedro trirrectángulo).</span><br />
<span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: 13px; line-height: 18px;">Para obtener la verdadera medida del cubo que se proyecta de forma ortogonal sobre el plano del cuadro, se abaten las caras, de manera que podamos trabajar sobre el papel, sobre el plano del cuadro PC.</span><br />
<span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: 13px; line-height: 18px;">Sobre la cara abatida (x) (y) se colocan las vistas de la pieza y se proyectan ortogonalmente sobre la traza de la cara abatida hasta que cortan a los ejes x’ y’, obteniendo así la dimensión reducida sobre los mismos de las aristas del cubo.</span><br />
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<a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj7YVw7GeO_8y9zclmwdJBWdUsZr9M60V9zzlDgZMw2HFxfouJjGTy_HExUkTynbyk4nrK3dvkDUJAqOxqGXe0GriddTkz17pv69HsQqQ1ucN94J9QYvBs4ZjDCBrU-lSGf4s_AB_t6wLST/s1600/4+axo+sd.jpg" style="background-color: white; color: #cc0000; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: 13px; line-height: 18px; text-decoration: none;"><img alt="" border="0" id="BLOGGER_PHOTO_ID_5599858894556459298" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEj7YVw7GeO_8y9zclmwdJBWdUsZr9M60V9zzlDgZMw2HFxfouJjGTy_HExUkTynbyk4nrK3dvkDUJAqOxqGXe0GriddTkz17pv69HsQqQ1ucN94J9QYvBs4ZjDCBrU-lSGf4s_AB_t6wLST/s320/4+axo+sd.jpg" style="-webkit-box-shadow: rgba(0, 0, 0, 0.0976563) 1px 1px 5px; background-attachment: initial; background-clip: initial; background-image: initial; background-origin: initial; border-bottom-color: rgb(238, 238, 238); border-bottom-style: solid; border-bottom-width: 1px; border-color: initial; border-image: initial; border-left-color: rgb(238, 238, 238); border-left-style: solid; border-left-width: 1px; border-right-color: rgb(238, 238, 238); border-right-style: solid; border-right-width: 1px; border-top-color: rgb(238, 238, 238); border-top-style: solid; border-top-width: 1px; border-width: initial; box-shadow: rgba(0, 0, 0, 0.0976563) 1px 1px 5px; float: left; height: 159px; margin-bottom: 10px; margin-left: 0px; margin-right: 10px; margin-top: 0px; padding-bottom: 5px; padding-left: 5px; padding-right: 5px; padding-top: 5px; position: relative; width: 320px;" /></a><br />
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<span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: 13px; line-height: 18px;">En el dibujo tenemos la representación en <a href="http://sistema-diedrico.blogspot.com.es/">sistema diédrico</a> del triedro al que se ha abatido una cara, el abatimiento lo observamos en el alzado mediante el giro de la cara xy. Sobre la cara abatida (x) (y) se coloca una de las caras de la figura y se proyecta mediante ortogonales a la charnela hasta que intercepta a los ejes xy.</span><br />
<span style="background-color: white; color: #222222; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: 13px; line-height: 18px;">Al proyectar la forma plana tenemos que los vértices de esta inciden sobre los ejes de la axonometría xy, con lo que tenemos ya la perspectiva de la cara de la figura con su reducción correspondiente y en perspectiva axonométrica.</span><br />
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<a href="http://perspectiva-axonometrica.blogspot.com/" style="background-color: white; color: #cc0000; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: 13px; line-height: 18px; text-decoration: none;">http://perspectiva-axonometrica.blogspot.com/</a><a href="http://perspectiva-axonometrica.blogspot.com/" style="background-color: white; color: #cc0000; font-family: Arial, Tahoma, Helvetica, FreeSans, sans-serif; font-size: 13px; line-height: 18px; text-decoration: none;"></a><br />
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<b>Dadas 3 líneas AB DB CB en el plano no coincidentes e incidentes en un punto B, determinar el triedro trirrectángulo del espacio que puede transformarse en la tres líneas por proyección.</b><br />
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Dados los tres ejes (en verde), construimos el triángulo cuyos ejes son las alturas y considerando éste como la base de la pirámide cuyas aristas son los 3 ejes en el espacio, abatimos sus caras (en azul), obteniendo así la verdadera forma de las mismas:<br />
<title>Axonométrico-fundamento - GeoGebra Hoja Dinámica</title><br />
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<table border="0" style="width: 600px;"><tbody>
<tr><td><br />
<h2>Axonométrico-fundamento</h2><br />
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</tbody></table>Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com1tag:blogger.com,1999:blog-4787394509272311049.post-62668598637566769082012-03-18T15:26:00.002-07:002012-03-18T15:31:19.609-07:00Teorema de Brahmagupta<title>Teorema de Brahmagupta - GeoGebra Hoja Dinámica</title><br />
<span style="background-color: white; font-family: Georgia, serif; line-height: 19px;">Si las diagonales de un cuadrilátero inscrito en una circunferencia son perpendiculares, entonces toda recta perpendicular a un lado cualquiera del cuadrilátero incidente en la intersección de las diagonales, divide el lado opuesto en dos partes iguales.</span><br />
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<table border="0" style="width: 600px;"><tbody>
<tr><td>En la figura siguiente: DG=GE=GC<br />
<h2>Teorema de Brahmagupta</h2><br />
<br />
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</applet></td></tr>
</tbody></table>Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com0tag:blogger.com,1999:blog-4787394509272311049.post-22939169982216345402012-03-18T14:57:00.003-07:002012-03-20T00:35:22.473-07:00Teorema de reciprocidad polar<title>Teorema de reciprocidad polar - GeoGebra Hoja Dinámica</title><br />
<br />
<span style="font-size: large;">Todas las polares de un punto B sobre una recta exterior a la circunferencia pasan por el mismo punto I, la recíproca es cierta.</span><br />
<br />
<table border="0" style="width: 600px;"><tbody>
<tr><td>Mover el punto B para observar que todas las polares FH pasan por el punto I :<br />
<h2>Teorema de reciprocidad polar </h2><br />
<br />
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</applet></td></tr>
</tbody></table>Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com0tag:blogger.com,1999:blog-4787394509272311049.post-1429575638846106972012-03-14T12:20:00.001-07:002012-03-14T12:23:33.733-07:00Teorema de Chasles<title>Teorema de Chasles - GeoGebra Hoja Dinámica</title><br />
<br />
<br />
<style type="text/css">
<!--body { font-family:Arial,Helvetica,sans-serif; margin-left:40px }-->
</style><br />
Haciendo un único giro sobre un plano se puede transformar una figura en otra igual, sea cual sea la posición de ambas.<br />
Para conseguir el centro de giro que transforma una figura en otra se une un punto con su transformado mediante un segmento, a continuación se hace lo mismo con otro par de puntos y en la intersección de las mediatrices de ambos segmentos tenemos el centro de giro.<br />
<br />
<table border="0" style="width: 600px;"><tbody>
<tr><td><a href="http://giros-traslaciones-simetrias.blogspot.com/">http://giros-traslaciones-simetrias.blogspot.com/</a><br />
<br />
<br />
<h2>Teorema de Chasles</h2><br />
<br />
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</applet></td></tr>
</tbody></table>Dr. Néstor Martín Gulias, catedrático de dibujo técnicohttp://www.blogger.com/profile/10824608066079221490noreply@blogger.com0