Let $ABCD$ be a cyclic quadrangle, let the diagonals $AC$ and $BD$ cross at $O$, and let $I$ and $J$ be the incentres of the triangles $ABC$ and $ABD$, respectively. The line $IJ$ crosses the segments $OA$ and $OB$ at $M$ and $N$, respectively. Prove that the triangle $OMN$ is isosceles.
Problem
Source: Danube 2015 Seniors P1
Tags: geometry, isosceles, cyclic quadrilateral, incenter
Andrei246
30.06.2020 15:50
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Let $ E $ be the intersection of sides $ DI$ and $CJ$. Claim Quadrilater $ ABIJ$ is cyclic. Proof Straightforward from incenter lemma we obtain $EA=EJ=EI=EB$. Thus $ E $ is the centre of circle $(AIBJ)$ Let $ \angle ADJ=x$ ,$\angle BAJ=y$ and $\angle IBC=z$ . Now from simple angle chasing we get $ \angle ONM=\angle OMN=180^{\circ}-2x-y-z $,which leads to the conclusion.
dgreenb801
02.07.2020 03:35
See my solution on my Youtube channel here: https://www.youtube.com/watch?v=rqUOlzDS_Nw
almagest3001
15.12.2024 14:15
Let $K_1, K_2$ be the incenters of triangles $\Delta AOB, \Delta DCO$; $L_1, L_2$ be the arc midpoints of $AB$ and $CD$. $K_1 L_1 K_2 L_2$ is a kite, hence $K_1K_2 \perp L_1L_2$. By Pascal this gives $K_1O \perp L_1L_2$. Using the fact that $ABIJ$ is cyclic and some angle chasing, $L_1L_2$ is parallel to $IJ$, so $K_1O \perp IJ$. We are done now, since a triangle is isosceles if and only if the angle bisector opposite to a side is also an altitude.
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