Question

已知ABCD是正方形,P是BC上一点,Q是CD上一点,PQ＝BP＋DQ.求角PAQ的度数.

Answer 1

延长CD至H,使DH=BP,连AH.

AB=AD ⇒RT△ABP≅RT△ADH

⇒AP=AH

BP+DQ=PQ

⇒DH+DQ=PQ

HQ=PQ

AQ=AQ

⇒△APQ≅△AHQ

∠PAQ=∠HAQ=∠BAP+∠DAQ=∠BAD/2=45°