Let ∠DBQ=x°\angle DBQ=x\degree∠DBQ=x°, so ∠PBD=60°−x°\angle PBD=60\degree-x\degree∠PBD=60°−x°.
∠ABP=60°−∠PBD=60°−(60°−x°)=x°.\angle ABP=60\degree-\angle PBD=60\degree-(60\degree-x\degree)=x\degree.∠ABP=60°−∠PBD=60°−(60°−x°)=x°.
If ∠ABP=∠DBQ\angle ABP=\angle DBQ∠ABP=∠DBQ, ∠BAP=∠BDQ\angle BAP=\angle BDQ∠BAP=∠BDQ, AB=BDAB=BDAB=BD, so △ABP=△DQB\triangle ABP=\triangle DQB△ABP=△DQB
and BP=BQBP=BQBP=BQ.
In △PBQ:∠PBQ=60°,PB=BQ\triangle PBQ: \angle PBQ=60\degree, PB=BQ△PBQ:∠PBQ=60°,PB=BQ, so ∠QPB=∠PQB=180°−∠PBQ2=60°\angle QPB=\angle PQB=\frac{180\degree-\angle PBQ}{2}=60\degree∠QPB=∠PQB=2180°−∠PBQ=60°.
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