证明:
(1)
∵CE⊥AB, BD⊥AC, ∴ ∠BEC=∠CDB=90°
∵在Rt△ABD和Rt△AEC中,∠BAC+∠ABD=∠BAC+∠ACE=90°
∴∠ABD=∠ACE
∠在△ABP和△QCA中,
AB=CQ, ∠ABD=∠ACE, PB=AC,
∴△ABP全等于△QCA, ∴ AP=AQ.
(2)
∵△ABP全等于△QCA, ∴ ∠APB=∠QAC, 即∠APD=∠QAC
∵ BD⊥AC, ∴ ∠ADP=90°
又∵在△ADP中,∠DAP+∠APD=90°
∴ ∠QAP=∠DAP+∠QAC=90°, ∴AP⊥AQ