Brazilian Math Olympiad

Draw $P{P}^{\prime }$ parallel to $IA$ so that $P^{\prime }$ is on line $AB$. Then $△PA{P}^{\prime }$ is isosceles, which implies that $BC=AB+AP=AB+A{P}^{\prime }=B{P}^{\prime }$. This then implies that $△P^{\prime }BC$ is isosceles, which in turn implies that, since $P$ is on the angle bisector of $\angle B$, $P^{\prime }PC$ is also isosceles, with $P{P}^{\prime }=PC$. It then follows, using similarity of triangles and the angle bisector theorem, that $$\frac{IA}{P{P}^{\prime }}=\frac{BA}{B{P}^{\prime }}=\frac{BA}{BC}=\frac{AP}{PC}=\frac{AP}{P{P}^{\prime }}$$ from which $IA=AP$.