如图,在△ABC中,∠ACB=90°,ae=bd,d为ac上的点,延长bc到点e,使ce=cd.求证:bd⊥ae
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![如图,在△ABC中,∠ACB=90°,ae=bd,d为ac上的点,延长bc到点e,使ce=cd.求证:bd⊥ae](/uploads/image/z/6872502-30-2.jpg?t=%E5%A6%82%E5%9B%BE%2C%E5%9C%A8%E2%96%B3ABC%E4%B8%AD%2C%E2%88%A0ACB%3D90%C2%B0%2Cae%3Dbd%2Cd%E4%B8%BAac%E4%B8%8A%E7%9A%84%E7%82%B9%2C%E5%BB%B6%E9%95%BFbc%E5%88%B0%E7%82%B9e%2C%E4%BD%BFce%3Dcd.%E6%B1%82%E8%AF%81%EF%BC%9Abd%E2%8A%A5ae)
如图,在△ABC中,∠ACB=90°,ae=bd,d为ac上的点,延长bc到点e,使ce=cd.求证:bd⊥ae
如图,在△ABC中,∠ACB=90°,ae=bd,d为ac上的点,延长bc到点e,使ce=cd.求证:bd⊥ae
如图,在△ABC中,∠ACB=90°,ae=bd,d为ac上的点,延长bc到点e,使ce=cd.求证:bd⊥ae
△AEC全等于△BDC
∠EAC=∠DBC
延长BD交AE于F
∠FAD=180°-∠FDA-∠EAC=180°-∠BDC-∠DBC=∠ACB=90°
所以BD⊥AE