P是正方形ABCD中DC边上一点,角BAP的角平分线交BC于Q.求证：AP=DP+BQ

Question

P是正方形ABCD中DC边上一点,角BAP的角平分线交BC于Q.求证：AP=DP+BQ

Answer 1

延长CD至E使DE=BQ
易知△ABQ≌△ADE
∠AQB=∠E=∠DAQ
∵∠AQB=∠DAQ=∠DAP+∠PAQ
且∠EAD=∠PAQ=∠QAB
∴∠DAP+∠PAQ=∠DAP+∠EAD
∠DAQ=∠EAP=∠E
∴AP=EP
AP=DP+BQ