已知△ABC是等腰直角三角形,∠BAC=90°,点D是BC的中点,作正方形DEFG,使点A,C分别在DG和DE上,连接AE,BG(1)试猜想线段BG和AE的数量关系(2)将正方形DEFG绕点D逆时针方向旋转一定角度后(大于0°,

来源:学生作业帮助网 编辑:作业帮 时间:2024/05/03 22:26:50
已知△ABC是等腰直角三角形,∠BAC=90°,点D是BC的中点,作正方形DEFG,使点A,C分别在DG和DE上,连接AE,BG(1)试猜想线段BG和AE的数量关系(2)将正方形DEFG绕点D逆时针方向旋转一定角度后(大于0°,

已知△ABC是等腰直角三角形,∠BAC=90°,点D是BC的中点,作正方形DEFG,使点A,C分别在DG和DE上,连接AE,BG(1)试猜想线段BG和AE的数量关系(2)将正方形DEFG绕点D逆时针方向旋转一定角度后(大于0°,
已知△ABC是等腰直角三角形,∠BAC=90°,点D是BC的中点,作正方形DEFG,使点A,C分别在DG和DE上,连接AE,BG
(1)试猜想线段BG和AE的数量关系
(2)将正方形DEFG绕点D逆时针方向旋转一定角度后(大于0°,小于360°)判断(1)中的结论是否仍然成立?如果成立,证明;不成立说明理由
(3)若BC=DE=2,在(2)的旋转过程中,当AE为最大值时,求AF的值

已知△ABC是等腰直角三角形,∠BAC=90°,点D是BC的中点,作正方形DEFG,使点A,C分别在DG和DE上,连接AE,BG(1)试猜想线段BG和AE的数量关系(2)将正方形DEFG绕点D逆时针方向旋转一定角度后(大于0°,
答案在图片中啦,只有10分太少了
那么长

相关的主题文章:



MBT Kimondo Shoes home,PUMA Voltaic Shoes, selling shoes, MBT Chapa MBT shoes, he began designing and Miller explore the comments, often in the holidays to the East Af...

全部展开

相关的主题文章:



MBT Kimondo Shoes home,PUMA Voltaic Shoes, selling shoes, MBT Chapa MBT shoes, he began designing and Miller explore the comments, often in the holidays to the East African savannah in East Africa barefoot in the absence of strict requirements, MBT Sand Shoe shoes discount brokers heelchapa spineless distance easily on the surface, our problems, he returned to his love of sales is through his endless inspiration from the back of torture is almost non-existent in this area. MBT sandals sale inference,Lebron James Shoes, back pain and arthritis are not present in the war Tankeshapa GTX mores.MBT owner, and from Switzerland to convince Kaermule,PUMA Drift Cat Shoes, Main Battle Tank Sang shoes is due to urbanization Construction of the foot.
MBT kaya shoes Main Battle Tank is the principle of participation m.walk cheap shot MBT shoes, a new main tanks storm GTX boots, shoes and the main battle Tankejimeng many things to bring the principles of reducing the portion of the shell line boots, and some extent differences feel.these boots all the principles involved in other ways Tankelami shoes boots your earnings to the principles of mbt shoes resistance to keep the load of the shell better tchanga birch, cheap sports shoes a better position to support major MBT battle to Section mystery, walk on the beach, perfect for the fun.
MBT online shop sales of the main tank, you can go to a parallel project, can help you decide what to buy more brands. Another advantage is the online store a large pool of low prices mbt fisherman sandal to save the store more than clothes. You may not know that any brand extension, that the pink shopping branch projects. More options, you can add), you and your favorite brand to buy more in-depth cab fisherman sandals, your wardrobe.
mbt panda sandals collection of available analysis if the absolute is cost, accumulated to read.

收起

已知,如图,△ABC是直角三角形,∠ABC=15°,∠BAC=90°,以BC为斜边作等腰直角三角形BCD,CD、BA交于点E △ABC是等腰直角三角形,∠BAC=90°,be是角平线,ed⊥bc,证ad垂直be 已知如图△ABC是等腰直角三角形,∠BAC=90°过点C作BC的垂线l,把一个足够大的三角板的直角顶点放到点A处8.已知:如图,△ABC是等腰直角三角形,∠BAC=90°,过点C作BC的垂线l,把一个足够大的 如图1已知三角形ABC与三角形ADE是等腰直角三角形角BAC=角DAE=90度 求解数奥题,已知△ABC是等腰直角三角形,∠BAC=90°,点D是△ABC内一点,且∠DAC=∠DCA=15°,求证:BD=BA 1.已知在△ABC中,AB=AC,D是BC上一点,E是线段AD上一点,且∠BED=2∠CED=∠BAC.求证:BD=2CD.2.已知△ABC是等腰直角三角形,∠BAC=90°,点D是△ABC内一点,且∠DAC=∠DCA=15°.求证:BD=BA.3.等腰直角三角形ABC中,延 已知△ABC是等腰直角三角形.∠BAC=90°,BD平分∠ABC,CE⊥BD交BD延长线于E.求证:BD=2CE.如图: △ABC是等腰直角三角形 ∠BAC=90° D是ABC内一点 ∠DAC=∠DCA=15° 求证BD=BA 如图,已知三角形abc是等腰直角三角形,∠bac=90°,点d是bc中点,做正方形defg连接ae如图,已知三角形ABC是等腰直角三角形,∠BAC=90°,点D是BC中点,做正方形DEFG,连接AE,若BC=DE=2,将正方形DEFG绕D逆时针旋 一道初三数学竞赛题已知△ABC和△ADE是等腰直角三角形,∠BAC=∠DAE=90°,点M是BE中点,求证:AM⊥DC.能说细点吗 已知:如图,在△ABC中,∠BAC=90°,AB=AC,D是BC上一点.EC⊥BC,且CE=BD.求证△ADE是等腰直角三角形. 已知:如图,在△ABC中,∠BAC=90°,AB=AC,D是BC上一点.EC⊥BC,且CE=BD.求证△ADE是等腰直角三角形. 如图已知在三角形abc中,∠bac=90°,ab=ac,d是bc上一点ec⊥bc,且ce=bd,求证△ade是等腰直角三角形急, 一道初三几何题,如图,△ABC是等腰直角三角形,∠BAC=90°,D是BC中点,三角形EFD也是等腰直角三角形请问能证明AD≠EF吗?望高手指教, 如图①,已知△ABC是等腰直角三角形,∠BAC=90°,点D是BC的中点.作正方形DEFG,使点A如图①,已知△ABC是等腰直角三角形,∠BAC=90°,点D是BC的中点.作正方形DEFG,使点A,C分别在DG和DE上,连接AE,BG.(1) 已知:如图5-5,Rt△ABC中,∠BAC=90°,AB=AC,D是BC的中点,AE=BF.求证:(1)DE=DF(2)△DEF为等腰直角三角形 一道关于勾股定理的数学题已知△ABC为等腰直角三角形,∠BAC=90°,E、F是BC边上的点,且∠EAF=45°,求证:BE²+CF²=EF² △ABC和△ADE都是等腰直角三角形,∠BAC=∠DAE=90°,点M是BE中点,求证:AM⊥CD