Parallel through the incenter in a right triangle

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In triangle $A\hat{B}C$, right-angled at $\hat{C}$, draw the bisectors of the interior angles $\hat{A}$ and $\hat{B}$. They meet at a point $I$. Draw through $I$ the parallel to side $BC$, meeting side $AC$ at $K$ and the hypotenuse $AB$ at $R$. Prove that $IK:IR=AC:AB$.

Solution

$I$ is the incenter (intersection of the bisectors). Put the right-angle vertex at the origin with the legs on the axes: $C(0;0),\qquad A(q;0)\ \ (q=\overline{AC}),\qquad B(0;p)\ \ (p=\overline{BC}).$ The incenter of a right triangle with legs on the axes has coordinates $I(r;r)$, where $r=\dfrac{p+q-h}{2}$ is the inradius and $h=\overline{AB}=\sqrt{p^2+q^2}$. The parallel to $BC$ (the $y$-axis) through $I$ is the vertical line $x=r$; it meets $AC$ ($x$-axis) at $K(r;0)$ and the hypotenuse at $R(r;y_R)$. So $IK=r,\qquad IR=y_R-r.$ The hypotenuse $AB$: $\dfrac{x}{q}+\dfrac{y}{p}=1\Rightarrow y_R=p\!\left(1-\dfrac{r}{q}\right)$, whence $IR=\frac{pq-r(p+q)}{q}.$ Now, using $r=\dfrac{p+q-h}{2}$: $pq-r(p+q)=\frac{-h^2+(p+q)h}{2}=\frac{h(p+q-h)}{2}=h\,r.$ Hence $IR=\dfrac{h\,r}{q}$ and $\frac{IK}{IR}=\frac{r}{\,h r/q\,}=\frac{q}{h}=\frac{\overline{AC}}{\overline{AB}}.$ $\boxed{IK:IR=\overline{AC}:\overline{AB}.}\qquad\blacksquare$