Problem 1

Let

ABC

be a triangle with incenter

I

. A point

P

in the interior of the triangle satisfies

\angle PBA+\angle PCA = \angle PBC+\angle PCB

. Show that

AP \geq AI

, and that equality holds if and only if

P=I

Step 3 of 4: Use the incenter-excenter center

M

on ray

AI

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In plain words

By the incenter-excenter lemma, the circumcircle of $\triangle BIC$ is centered at the midpoint $M$ of arc $BC$ (where ray $AI$ meets the circumcircle of $\triangle ABC$ ), so $P$ and $I$ are at the same distance from $M$ .

MB=MC=MI=MP,\quad AM=AI+IM

Detailed analysis

Let ray $AI$ meet the circumcircle of $\triangle ABC$ again at $M$ , the midpoint of arc $BC$ not containing $A$ . Because $I$ lies on the segment from $A$ to $M$ , we have $AM=AI+IM$ . By the incenter-excenter lemma, $M$ is the circumcenter of $\triangle BIC$ , so $MB=MC=MI$ ; since $P$ also lies on that circle (Step 2), $MP=MI$ as well.