BP=AC AB=CQ ∠ABP=90 -∠BAC=∠QCA

So △ABP congruence △QCA So AP = AQ∠BPA =∠QAC.

∠ QAP =∠ QAC +∠ Cap =∠ BPA+∠ Cap =180-∠ ADP = 90 Note: Look at the right triangle ADP.

(2) similarity proof △AEC congruence △BDC

So ∠EAC=∠CBD, so

∠AFD = 180-∠FAD-∠ADF = 180-∠EAC-∠CDB = 180-∠CBD-∠CDB =∠BCD = 90

So AF⊥BD