Prove:
Extend QD to p' so that DP'=BP, even AP'
Then △ ADP '△ ABP
AP'=AP
∠P'AD=∠PAB
∠P' AQ =∠P' ad+∠DAQ =∠ Pabu +∠DAQ=90-∠PAQ=90-45=45.
So ∠P'AQ=∠PAQ.
So, △ AQP '△ AQP
AQP = AQP
S△ADQ+S△ADP'=S△APQ
S△ADQ+S△ABP=S△APQ