The diagonals of convex quadrilateral $ABCD$ intersect at point $E$. Triangles $ABE$ and $CED$ have a common excircle $\Omega$, tangent to segments $AE$ and $DE$ at points $B_1$ and $C_1$, respectively. Denote by $I$ and $J$ the centers of the incircles of these triangles, respectively. Segments $IC_1$ and $JB_1$ intersect at point $S$. It is known that $S$ lies on $\Omega$. Prove that the circumcircle of triangle $AED$ is tangent to $\Omega$. Proposed by David Brodsky
Problem
Source: 239 School Open MO, 2023, Junior league, Problem 7
Tags: geometry
mathuz
19.04.2023 22:08
very nice problem! Let $M$ be the midpoint of $B_1C_1$. Our aim will be showing that $M$ coincides with the incenter of the triangle $AED$. Indeed, this would be enough to say that circles $\Omega$ and $(AED)$ are tangent each to other (by the mixtilinear construction). It is clear that $IJ\parallel B_1C_1$ and $M$ lies on the internal bisector of $\angle AED$. Let $\varphi := \angle AED$. So, it suffices to show that $\angle AMD = 90^{\circ}+\frac{\varphi}{2}$. On the other hand, since $S$ lies on $\Omega$, we get $\angle B_1SC_1=90^{\circ}+\frac{\varphi}{2}$. Moreover, $\angle EIS=\angle SC_1B_1 = \angle SB_1E$ and $\angle EJS = \angle SB_1C_1 =\angle SC_1E$, and this gives us a family of similar triangles $\triangle B_1SC_1 \sim \triangle ISJ \sim \triangle IEC_1 \sim \angle JEB_1$. From this similarity, we obtain \[ EI\cdot EJ = EB_1^2=EC_1^2. \] We know that $AB_1 = EI\cdot \sin \frac{\varphi}{2}$ and $DC_1=EJ\sin \frac{\varphi}{2}$, so using the above result, \[ AB_1\cdot DC_1 = EI\cdot EJ \sin^2 \frac{\varphi}{2} = EC_1^2 \sin^2\frac{\varphi}{2} = C_1M^2. \]This implies the similarity of triangles $\triangle DC_1M$ and $\triangle AB_1M$ as $\angle AB_1M = \angle DC_1M = 90^{\circ} +\frac{\varphi}{2}$. Therefore, \[ \angle B_1AM=\angle C_1MD \quad \text{and} \quad \angle B_1MA = \angle C_1DM, \]which implies that $\angle AMD = \angle AB_1M = 90^{\circ}+\frac{\varphi}{2}$.