几何高手进...麻烦自己画图,不好意思啊如图,△ABC中,∠BAC=60°,∠ACB=40°,Q、P分别在在BC、AC上,并且AQ、BQ是∠BAC、∠ABC的角平分线,求证：BQ +AB=AP+BP

Question

几何高手进...麻烦自己画图,不好意思啊
如图,△ABC中,∠BAC=60°,∠ACB=40°,Q、P分别在在BC、AC上,并且AQ、BQ是∠BAC、∠ABC的角平分线,求证：BQ +AB=AP+BP

Answer 1

证明：
延长BP至D,使AP=DP；
AB与BQ交点为O
∠ABC=80
=>∠ABP=40
=>∠APD=100
=>∠PAD=∠ADP=40=∠ABD
=>AD=AB
∠CAQ=∠BAQ=30
=>∠QAD=70
∠APD=100,∠QAC=30
=>∠AOD=70=∠QAD
=>AD=OD=AB
∠BAQ=30,∠ABC=80
=>∠BQA=70=∠AOD=∠BOQ
=>BO=BQ
所以：
BQ+AB=BO+AB=BO+OD=BD=PD+BP=AP+BP
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