如图 abc和 Ade是有公共顶点的等腰直角三角形∠bac=∠dae=90°点p为射线bdce的交点．1求证bd=ce2若 Sketch Coloring Page

如图 abc和 Ade是有公共顶点的等腰直角三角形∠bac=∠dae=90°点p为射线bdce的交点．1求证bd=ce2若 Template

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tomski如图 abc和 ade是有公共顶点的等腰直角三角形∠bac=∠dae=90°点p为射线bdce的交点．1求证bd=ce2若 coloring

如图 ABC和 ADE是有公共顶点的等腰直角三角形∠BAC=∠DAE=90°点P为射线BDCE的交点．1求证BD=CE2若 sketch template

如图 ABC与 ADE都是以A为直角顶点的等腰直角三角形DE交AC于点F且AB=5AD=32．当 CEF是直角三角形时BD sketch template

如图 ABC和 ADE都是等腰直角三角形∠ACB=∠ADE=90°AC=2AD=1F为BE的中点．1如图1当边AD与边AB重合 sketch template

从一道简单的高数题谈起——点到三角形边及顶点的距离问题知乎 sketch template

题目已知 ABC和 ADE都是等腰直角三角形且∠BAC=∠DAE=90°1如图�点DE分别在线段ABAC上请直接写出线段 sketch template

问题�现如图� ABC与 ADE是等边三角形且点BDE在同一直线上连接CE求∠BEC的度数并确定线段BD与CE的数量关系．拓展 sketch template

如图 ABC和 ADE都是等腰三角形AB=ACAD=AE且∠BAC=90°∠DAE=90°BCD在同一条直线上．则线段BD与 sketch template

定义：两个顶角相等且顶角顶点重合的等腰三角形组合称为”相似等腰组”如图1，等腰 ABC和等腰 ADE即为“相似等腰组” sketch template

如图所示 ABC和 ADE都是等腰直角三角形且∠BAC=∠EAD=90°连接BDCE．1求证BD=CE2观察图形猜想BD sketch template

如图 ABC和 ADE均为等边三角形BDE三点在同一条直线上若∠ACE=20°则∠BAD= 度．题目和参考答案——青夏教育精英家教�—— sketch template

如图， ABC和 DEF是两个全等的等腰直角三角形，∠ BAC=∠ EDF=90°， DEF的顶点E与 ABC的斜边BC的中点重合，将 DEF sketch template

如图 ABC和 ADE为等腰直角三角形其中∠BAC=∠DAE=90°AB=ACAD=AE将 ADE绕A点旋转至如图的位置1判断 sketch template

如图 ABC是等腰三角形∠C=90°D是AB的中点点EF分别在ACBC边上运动且保持AE=CF连接DEDFEF．在此运动 sketch template

ABC和 ADE中AB=ACDA=DE∠BAC=∠ADE=α点D在BC上连接CE1如图1若α=60°则∠DCE=120 sketch template

26．如图在 ABC中已知AB = AC∠BAD =∠CAE点DE在BC上试说明 ADE是等腰三角形的理由．——青夏教育精英家教�—— sketch template

如图在 ABC中BDCE分别为三角形的两条高交于点O �问图11中有 个三角形与相似它们是 �连结DE ADE与 ABC是否相似 sketch template

初学讲义之高中数学二十：简单几何体知乎 sketch template

如图已知 ABC和 ADE都是等腰直角三角形∠BAC=∠DAE=90°AB=ACAD=AE．连接 BD交AE于M连接CE交AB于N sketch template

如图在 ABC中BDCE分别为三角形的两条高交于点O �问图11中有 个三角形与相似它们是 �连结DE ADE与 ABC是否相似 sketch template

如图在Rt ABC中∠BAC=90°AB=AC=2点D在BC所在的直线上运动作∠ADE=45°．若点D在线段BC上运动DE交AC于 sketch template

如图1在 ABC和 ADE中∠BAC=∠EADAB=ACAD=AE连接CDAE交于点F．1求证∠DCE=∠BAC2当∠ sketch template

如图在等腰 ABC中∠BAC=90°AB=AC=1点D是BC边上的一个动点在AC上取一点E使∠ADE=45°．1求证 ABD sketch template

（1）问题�现：如图1， ABC和 ADE都是等边三角形，连接BD，CE．易知=．（2）类比探究如图2， ADE都是直角三角形，∠ sketch template

【操作�现�（1）如图1，在 ABC和 ADE中，∠DAE=∠BAC，AD=AE，AB=AC．求证： ABD≌ ACE；【问题解决�（2）如图 sketch template

【中考提升�初中数学平面几何压轴题6大模型及解题方法知乎 sketch template

如图1在 ABC和 ADE中∠BAC=∠EADAB=ACAD=AE连接CDAE交于点F．1求证∠DCE=∠BAC2当∠ sketch template

历年中考数学几何压轴专题，中考几何专题训练及参考答案知乎 sketch template

初中几何最值专项1：定弦定角知乎 sketch template

如图已知 ABC和 ADE是等边三角形联结BDCE．1说明BD=CE的理由2延长BD交CE于点F求∠BFC的度数．题目和 sketch template

初中几何最值专项1：定弦定角知乎 sketch template

如图 ABC和 ADE都是等腰直角三角形∠BAC=∠DAE=90°连结CE交AD于点F连结BD交CE于点G连结BE．下列结论中正确 sketch template

�们知道含有36°的等腰三角形是特殊的三角形通常把有一个内角等于36°的三角形称为“黄�三角形 ．1如图12在 ABC中已知 sketch template

如图，Rt ABC中，∠C＝90°，AC＝BC，点D、E分别是边AB、AC上的点，把 ADE沿DE折叠，点A恰好落在BC上的点F处，若点F为 sketch template

如图，在 ABC中，点D是BC上一点，E是AD上一点，EB=EC，∠ABE=∠ACE。求证：AD⊥BC sketch template

已知 ABC的面积为S 1如图a将AB三等分D1D2是三等分点且D1E1∥D2E2∥BC求梯形D1E1E2D2 的面积 sketch template

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