Answer :

Given that ABC is a triangle.

BP and CP are internal bisector of B and C respectively


BQ and CQ are external bisector of B and C respectively.


External B = 180o - B


External C = 180o - C


In


BPC + B + C = 180o


BPC = 180o - (B + C) (i)


In


BQC + (180o - B) + (180o - C) = 180o


BQC + 180o - (B + C) = 180o


BPC + BQC = 180o [From (i)]


Hence, proved


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