A circumcircle of triangle $ABC$ meets the sides $AD$ and $CD$ of a parallelogram $ABCD$ at points $K$ and $L$ respectively. Let $M$ be the midpoint of arc $KL$ not containing $B$. Prove that $DM \perp AC$. by E.Bakaev
Problem
Source: Sharygin Geometry Olympiad 2016 Final Round problem 2 grade 8
Tags: geometry, arc midpoint, circumcircle, perpendicular, parallelogram
PROF65
22.07.2018 23:13
$\angle BAK =\angle LCB $ so $BK=BL$ thus $BM$ is a diameter hence $AM\perp AB\parallel DC, CM\perp CB\parallel DA$ then $DM\perp AC$
adiarasel
23.07.2018 02:28
that is BEAUTIFUL!!!!!
Plops
08.10.2019 02:12
Notice since $A,B,C,L,K$ are concylic, $\angle ABC= \pi-\angle ALC= \angle ALD$. Also, since $ABCD$ is a parallelogram, $\angle ABC=\angle ADC$, so $\angle ALD=\angle ADC=\angle ADL=\beta$. Let $\angle KML=\alpha \implies \angle KAl=\pi-\alpha$. By $\triangle ADL$, $\angle DAL+\angle ADL+\angle ALD=\pi-\alpha+2\beta=\pi \implies \alpha=2\beta$. Since, $M$ lies on the perpendicular bisector of $KL$, it follows that $M$ is the circumcenter of $\triangle DKL$. Since $KL$ is a transversal of $AC$, $M$ lies on the perpendicular from $D$ to $AC$.
OronSH
08.10.2019 03:56
Plops wrote: Notice since $A,B,C,L,K$ are concylic, $\angle ABC= \pi-\angle ALC= \angle ALD$. Also, since $ABCD$ is a parallelogram, $\angle ABC=\angle ADC$, so $\angle ALD=\angle ADC=\angle ADL=\beta$. Let $\angle KML=\alpha \implies \angle KAl=\pi-\alpha$. By $\triangle ADL$, $\angle DAL+\angle ADL+\angle ALD=\pi-\alpha+2\beta=\pi \implies \alpha=2\beta$. Since, $M$ lies on the perpendicular bisector of $KL$, it follows that $M$ is the circumcenter of $\triangle DKL$. Since $KL$ is a transversal of $AC$, $M$ lies on the perpendicular from $D$ to $AC$. How did you even find this thread?