if AP+AB = CB, then triangle API is isosceles, I incenter, BP angle bisector

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if AP+AB = CB, then triangle API is isosceles, I incenter, BP angle bisector

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#1 h Sep 28, 2018, 7:21 AM • 1 Y

Y by Adventure10

Let $I$ be the center of a circle inscribed in a triangle $ABC$ , $BP$ be the bisector of the angle $\angle ABC$ , $P$ lies on $AC$ . Prove that if $AP + AB = CB$ , then the triangle $API$ is isosceles.

This post has been edited 1 time. Last edited by parmenides51, Sep 28, 2018, 7:21 AM
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#2 h Sep 28, 2018, 8:24 AM • 2 Y

Y by Adventure10, Mango247

We note point $K$ on side $BC$ such $BA=BK$ . We have $PA=KC=PA$ , and $\angle AIP=180-\angle AIB=90-\frac{1}{2}\angle BCA$ , and $\angle API = 180-\angle BPC=180-\angle BPK-\angle KPC$ $=180-(90-\frac{1}{2}\angle KPC)-\angle KPC=90-\frac{1}{2}\angle BCA$ . We have that $AI=AP$ We are done!

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