Question

在Rt△ABC中,角C=90°,D,E分别是AC,BC边上的点,且BE=AC,EC=AD,连接AE、BD相交于点P.求角

Answer 1

证明：过E做ME⊥BC,是ME = CE.连结MD,MB∵∠C = 90°∴AC//ME∵AD = CE∴AD = ME∴四边形AEMD是平行四边形∴MD = EA,∠DME = ∠DAE∵在△MEB和△ECA中ME = CE,∠MEB = ∠C,EB = AC∴△MEB≌△ECA∴MB = AE,∠BME = ∠CEA∴MD = MB∵在△ACE中,∠CAE + ∠CEA = 90°∴∠DME+∠BME = 90°∴△DMB是Rt等腰△∴∠MDB = 45°∵DM//AE∴∠EPB = ∠MDB = 45°