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


BPC + B + C = 180o

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


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

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

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

Hence, proved

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