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 1 of 4: Compute

\angle BPC

from the angle condition

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In plain words

The two angle sums $(\angle PBA+\angle PCA)$ and $(\angle PBC+\angle PCB)$ together make $\angle B+\angle C$ ; being equal, each is half of $\angle B+\angle C$ , which is also what the angle bisectors cut off for $I$ .

\angle PBC+\angle PCB=\frac{\angle B+\angle C}{2}\ \Longrightarrow\ \angle BPC=180^\circ-\frac{\angle B+\angle C}{2}=90^\circ+\frac{\angle A}{2}=\angle BIC

Detailed analysis

Since $P$ is inside $\triangle ABC$ , $\angle B=\angle PBA+\angle PBC$ and $\angle C=\angle PCA+\angle PCB$ . Adding these and using $\angle PBA+\angle PCA=\angle PBC+\angle PCB$ gives $2(\angle PBC+\angle PCB)=\angle B+\angle C$ , so $\angle PBC+\angle PCB=\frac12(\angle B+\angle C)$ . In $\triangle BPC$ this leaves $\angle BPC=180^\circ-\frac12(\angle B+\angle C)=90^\circ+\frac12\angle A$ , which is the standard value of $\angle BIC$ .