In triangle $ABC$ $\angle C=90^\circ$, $AA'$ and $BB'$ are angle bisectors intersecting at $I$ ($A'\in BC$, $B'\in AC$). What would be the easiest way to prove that projection of $I$ onto $AB$ lies in equal distances from projections of $A'$ and $B'$ onto $AB$?
The reason I’m asking about the easiest way rather than about any solution is that I think it’s quite clear all the lengths can be calculated either in terms of sides of triangle or trigonometrically via its angles. Both ways seem to be too ugly. So I’m hoping there is smart geometrical solution.
Draw the incircle (with center $I$ and radius $r$) of $\triangle ABC$, and let it meet $BC$ at $P$. Since $IP \perp BC$, we have $IP \parallel AC$. (Note: This is why the right angle at $C$ matters.) Also, $|IP|=r$.
Let $Q$ be the point "near" $A^\prime$ on the incircle such that $IQ \parallel AB$; then, $\angle PIQ = \angle A$ and $\angle A^\prime I Q = \angle A^\prime IP = \frac{1}{2}\angle A$. By SAS, $\triangle IPA^\prime \cong \triangle IQA^\prime$, whence $A^\prime Q \perp IQ$, which is to say: the radius $IQ$ is the projection of $IA^\prime$ onto a line parallel to $AB$. This projection is congruent to the projection of $IA^\prime$ onto $AB$ itself.
The same argument applies to the corresponding projection of $IB^\prime$, so that the projections of $A^\prime$ and $B^\prime$ onto $AB$ each lie at distance $r$ from the projection of $I$.