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In triangle ABC, the bisector of interior angle at vertex B and the bisector of exterior angle at vertex A intersect each other at point P. Prove that :

2angle APB = angle C.




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See the diagram. 

BP is the internal bisector of \angle{ABC}.  AP is the bisector of the exterior \angle{DAC} at A.

\angle{ABP}=\angle{PBC}=\frac{\angle{ABC}}{2},\\\angle{DAP}=\angle{PAC}=\frac{\angle{DAC}}{2}\\\\\angle{DAC}=exterior\ angle\ to\ \Delta ABC\ at\ A\\=sum\ of\ interior\ angles=\angle{ABC}+\angle{ACB}\\\\\angle{DAP}=exterior\ angle\ to\ \Delta PAB\ at\ A=\angle{APB}+\angle{ABP}\\\\

= > \frac{1}{2}\angle{DAC}=\angle{APB}+\frac{1}{2}\angle{ABC}\\= > \frac{1}{2}( \angle{ABC}+\angle{ACB}) =\angle{APB}+\frac{1}{2} \angle{ABC}\\

= >\frac{1}{2} \angle{ACB}=\angle{APB}\\\\= > \angle{C}=2*\angle{APB}

It is easily proved by using the exterior angle = sum of interior angles.

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