如图所示,在△ABC中,∠C=90°,∠F=∠HPD=∠DGA=90°.求证：△ABC相似于△DEF

证明：∵∠HPD＝∠F＝90°∴HP‖EF(同位角相等,两直线平行)∴∠DHP＝∠DEF(两直线平行,同位角相等)∵∠C＝∠DGA＝90°∴HG‖BC(同位角相等,两直线平行)∴∠AHG＝∠B(两直线平行,同位角相等)又∵∠AHG＝∠DHP(对顶角相等)∴∠DEF＝∠B,∠F＝∠C＝90°∴△ABC∽△DEF