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 2 of 4:

P

lies on the circumcircle of

\triangle BIC

Previous step Next step

\angle BPC=\angle BIC\ \Longrightarrow\ B,P,I,C\ \text{are concyclic}

Detailed analysis

Both $P$ and $I$ lie on the same side of line $BC$ (inside $\triangle ABC$ ), and by Step 1 they subtend segment $BC$ at equal angles $\angle BPC=\angle BIC$ . Therefore $P$ lies on the circumcircle of $\triangle BIC$ .