如图,已知ΔABC中,∠BAC=90º,四边形ABDE,BCFG是两个正方形,AB的延长线交DG于P,求证：A

过G做AC的平行线与BP的延长线交与M
∠M=∠BAC=90°
∠ABC+∠BCA=90°
∠ABC+∠PBG=90°
∠BCA =∠PBG
BG=BC
△BGM≌△AbC
AC=BM GM=AB=BD
△PMG≌△BDP
BP=1/2BM=1/2AC