A, D, E, F concyclic iff E'F '= M'N', 4 projections related

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A, D, E, F concyclic iff E'F '= M'N', 4 projections related

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Source: 2014 Thailand MO Shortlist G7 - TMO11

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#1 h Dec 13, 2020, 3:10 AM

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Let $ABC$ be an acute triangle with $E$ and $F$ on sides $AB$ and $AC$ respectively, and $O$ be it's circumcenter. Let $AO$ intersect $BC$ at point $D$ . Let the perpendicular from point $D$ on sides $AB$ and $AC$ intersect them at points $M,N$ respectively. Let the perpendicular on the side $BC$ from points $E,F,M,N$ intersect it at points $E',F',M',N'$ respectively. Prove that $A, D, E, F$ lie on the same circle if and only if $E'F '= M'N'$ .

This post has been edited 1 time. Last edited by parmenides51, Dec 13, 2020, 3:12 AM

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DNCT1

#2 h Dec 19, 2020, 4:54 PM • 1 Y

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$\textit{Proof:}\quad A,D,E,F\quad\text{concyclic}\implies E'F'=M'N'$
First we have $2$ claims
$\textbf{Claim 1}\$ : $\quad\boxed{MM'=NN'}$
$\textit{Proof}$
Let $AH$ be the altitude of $\bigtriangleup ABC$ with $H\in\overline{BC}$
By Thales it's suffice to show that $\frac{AM}{AN}=\frac{AB}{AC}$ which is
$\frac{\sin \angle ANM}{\sin \angle AMN}=\frac{\sin \angle ACB}{\sin \angle ABC}\iff \frac{\sin \angle ADM}{\sin \angle ADN}= \frac{\sin \angle ACB}{\sin \angle ABC}\quad(\text{which is true})$ since $\angle ADM=90^o-\angle BAO=\angle ACB$ and similarly $\angle ADN=\angle ABC.\blacksquare$
$\textbf{Claim 2:}\quad\boxed{\overline{ME'}\parallel\overline{NF'}}$
$\textit{Proof}$
Let $(AEF)\cap \overline{BC}=\{D,P\}$ since $E'MED$ concyclic we have
$\angle BE'M=\angle BED=180^o-\angle AED=180^o-\angle APD=\angle BPA\implies\overline{ME'}\parallel\overline{AP}$ Similarly we have $\overline{NF'}\parallel\overline{AP}$ which implies the claim holds. $\quad\blacksquare$
By $2$ Claims we have $\bigtriangleup MM'E=\bigtriangleup NN'F\implies M'E'=N'F'\implies\boxed{E'F'=M'N'}.\quad\blacksquare$
$\textit{Proof:}\quad E'F'=M'N'\implies A,D,E,F\quad\text{concyclic}$
We have $M'E'=N'F'\implies\overline{ME'}\parallel\overline{NF'}\implies \angle BE'M=\angle DF'N\implies \angle BED=\angle AFD\implies\boxed{A,D,E,F\quad\text{concyclic}}.\quad\blacksquare$

This post has been edited 1 time. Last edited by DNCT1, Dec 19, 2020, 6:05 PM

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