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