Problem 20: Concyclic points

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Joined: Mar 10, 2015

by henderson, Jun 7, 2016, 8:45 PM

$\bf\color{red}Problem\ 20\$ Let $ABC$ be an isosceles triangle such that $AB=AC$ and let $M$ be the midpoint of $BC$ . Consider a point $D$ on the segment $BC$ . Let $I$ and $J$ be the incenters of $\triangle ABD$ and $\triangle ACD,$ respectively. Prove that $M, I, J$ and $D$ are concyclic.

$\bf\color{red}Solution\$ WLOG assume that $BD < CD$ . Then denote by $P$ the intersection of $DJ$ and $AM$ .

$\boxed{1}$ $\angle{IAD} = \angle{JAM}$ .
Proof: We know that $\angle{IAJ}=\angle{IAD} + \angle{JAD} = \tfrac{1}{2}\angle{A}$ . Also $\angle{BAM} = \tfrac{1}{2}\angle{A},$ so consequently $\angle{IAD}=\angle{BAI}= \angle{JAM}.$
Since $I$ and $J$ are both incenters and $\angle{B} = \angle{C}$ , $\angle{AID} = \angle{AJP}$ . So, we deduce that $\triangle{AID} \sim \triangle{AJP}.$

$\boxed{2}$ $\triangle{IDM} \sim \triangle{JPM}.$
Proof: Firstly, $\angle{IDM} = 180^\circ - \angle{IDB} = 180^\circ - \angle{IDA} = 180^\circ - \angle{APJ} = \angle{JPM}$ . Also, by the angle-bisector theorem $,$ $\frac{DM}{MP} = \frac{DA}{PA}.$ But by similar $\triangle AID$ and $\triangle AJP$ , $\frac{DA}{PA} = \frac{ID}{PJ}$ , so we have $\frac{DM}{ID}=\frac{MP}{PJ}.$ Combining this fact with
$\angle IDM=\angle JPM,$ we conclude that $\triangle IDM \sim \triangle JPM.$ So, $\angle DIM=\angle PJM=\angle DJM,$ which means that $M, I, J, D$ are concyclic. $\square$

$\blacktriangleright$ Link to the problem
$\blacktriangleright$ Generalization of the problem

This post has been edited 12 times. Last edited by henderson, Jan 13, 2017, 10:32 AM

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Problem 20: Concyclic points

by henderson, Jun 7, 2016, 8:45 PM

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