如图,在△ABC中,∠B=∠C,D为BC边上一点,DF⊥BC交AC于点F,DE⊥AB交AB于点E.证明∠AFD=∠EDF+90°
来源:学生作业帮助网 编辑:作业帮 时间:2024/07/21 19:19:18
![如图,在△ABC中,∠B=∠C,D为BC边上一点,DF⊥BC交AC于点F,DE⊥AB交AB于点E.证明∠AFD=∠EDF+90°](/uploads/image/z/2561057-17-7.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2C%E2%88%A0B%3D%E2%88%A0C%2CD%E4%B8%BABC%E8%BE%B9%E4%B8%8A%E4%B8%80%E7%82%B9%2CDF%E2%8A%A5BC%E4%BA%A4AC%E4%BA%8E%E7%82%B9F%2CDE%E2%8A%A5AB%E4%BA%A4AB%E4%BA%8E%E7%82%B9E.%E8%AF%81%E6%98%8E%E2%88%A0AFD%3D%E2%88%A0EDF%2B90%C2%B0)
xR]oP+
ɮVҿauWl&2Mj2ИXf0ğBzJw_]$yy'eqF4m|r=ǵCUJ0{V';}Z-
2pj<2AwoJFSgbYU[o/el*nSS\NNJO[f:VT,[`
EPʌY`_^b_V5
Hf2㼀/
:x>aHX3tq%y!H"lbqlyKiqE3mS1>˒Xd!d`ĉ:F0Ĝ)&8/1RC?