Let $ABCD$ be a cyclic quadrilateral and let its diagonals $AC$ and $BD$ cross at $X$. Let $I$ be the incenter of $XBC$, and let $J$ be the center of the circle tangent to the side $BC$ and the extensions of sides $AB$ and $DC$ beyond $B$ and $C$. Prove that the line $IJ$ bisects the arc $BC$ of circle $ABCD$, not containing the vertices $A$ and $D$ of the quadrilateral.
Problem
Source: Romanian 2018 TST Problem 1 day 3
Tags: cyclic quadrilateral, incenter, geometry
HamstPan38825
05.09.2022 07:18
Let $\alpha, \beta, \gamma, \theta$ denote half the measures of $\widehat{AB}, \widehat{AD}, \widehat{DC}, \widehat{CB}$ respectively, and let $M$ be the midpoint of minor arc $\widehat{BC}$. Furthermore, let $p = \overline{AB} \cap \overline{CD}$, so $J$ is the $P$-excenter of $\triangle PBC$. It suffices to show that $$[ICM] + [CMJ] = [ICJ] \iff \frac{\sin \angle MCJ}{CI} + \frac{\sin \angle MCI}{CJ} = \frac{\sin \angle ICJ}{CM}.$$The rest is a (surprisingly short) computation. By the Law of Sines and some angle chasing we may rewrite \begin{align*} CI&= BC \cdot \frac{\sin \frac{\gamma}2}{\cos \frac{\beta+\theta}2} \\ CJ &= BC \cdot \frac{\cos \frac{\gamma+\beta}2}{\cos \frac{\theta-\beta}2} \\ CM &= \frac{BC}{2 \cos \frac{\theta}2}, \end{align*}and we also have \begin{align*} \sin \angle MCJ &= \sin \frac{\gamma}2 \\ \sin \angle MCI &= \sin \frac{\theta+\alpha}2 \\ \sin \angle ICJ &= \cos \frac{\beta}2. \end{align*}Now notice that $\cos \frac{\beta+\gamma}2 = \sin \frac{\alpha+\theta}2.$ Then the relation is equivalent to $$\cos \frac{\theta+\beta}2 + \cos \frac{\theta-\beta}2 = 2 \cos \frac{\theta}2 \cos \frac{\gamma}2$$after much cancellation, which is obvious by sum to product.