Let $ABC$ be a non-isosceles triangle with incenter $I$. Let $D$ be a point on the segment $BC$ such that the circumcircle of $BID$ intersects the segment $AB$ at $E\neq B$, and the circumcircle of $CID$ intersects the segment $AC$ at $F\neq C$. The circumcircle of $DEF$ intersects $AB$ and $AC$ at the second points $M$ and $N$ respectively. Let $P$ be the point of intersection of $IB$ and $DE$, and let $Q$ be the point of intersection of $IC$ and $DF$. Prove that the three lines $EN, FM$ and $PQ$ are parallel. Proposed by Saudi Arabia
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Tags: geometry, incenter
Aryan-23
12.09.2020 22:30
Cool angle chasing exercise. Firstly note that $I$ is the midpoint of arc $\widehat {DE}$ not containing $B$. So $ID=IE=IF$,implying $I$ is the circumcenter of $\odot (DEF)$. Note that : $$\angle PDQ = \angle PDI + \angle QID = \angle PBE + \angle QCF = \frac 12 B + \frac 12 C = \pi- \angle BIC \implies \{PIQD\} \quad \text {is cyclic}$$ This means that $\angle IPQ = \angle IDQ = \frac 12 C$. This combined with $\angle AIB = 90^{\circ}+ \frac 12 C $ implies $AI \perp PQ$. Now note that by Miquel $AEIF$ is cyclic. Assume that $A-M-E$ and $A-F-N$ are collinear in that order. Then, $\angle IEM = \angle IFN$. So we get that $\Delta IEM \cong \Delta IFN$ , which gives $ME=FN \implies MF \parallel FN$. Note that this also means that $AI \perp MF$ , so we are done $\blacksquare$
Com10atorics
23.09.2020 18:35
$\angle IDF=\tfrac {\angle C}{2}=\angle IFD$ so $I$ is the incenter of $(DEF)$. $\angle IMA=\angle IME=\angle IEB=\angle IDC=\angle IFN=\angle INF$ so $ANIM$ is cyclic. It is clear that $DINC$ is a deltoid. $\angle CIN=\angle CID=\angle CFD$ so $NIMQ$ is cyclic and similarly so is $MIEP$. Now $\angle DQI=\angle INF=\angle EMI=\angle EPI$ which means $DQIP$ is cyclic and finally, $\angle DQP=\angle DIB=\angle DEB=\angle DEM=\angle DFM$ So $QP||FM$ and similarly $QP||EN$