In acute triangle $ABC,$ $AB<AC,$ $O$ is the circumcenter of the triangle. $M$ is the midpoint of segment $BC,$ $(AOM)$ intersects the line $AB$ again at $D$ and intersects the segment $AC$ at $E.$ Prove that $DM=EC.$
Problem
Source: CWMI 2018 Q5
Tags: geometry, circumcircle
16.08.2018 14:13
SHOW THAT $EMC = DBM$ ITS VERY EASY Q.E.D
16.08.2018 21:09
Because $OM\bot BC, BC$ passes through $F|OF$ is a diameter of the circle $\odot (AOM)$, i.e. $AF$ is tangent to the circle $\odot (ABC)$, thus $\angle DMF=\angle DAF=\angle ACB$; with $\angle CEM=\angle BDM$ and $BM=CM\implies \triangle BDM\cong\triangle MCE$ (a.s.a. criterion). Best regards, sunken rock
07.07.2019 05:49
Let $U$ $\in$ $BC$ which satisfies $AU$ tangents $(ABC)$ at $A$ We have: $DM = \dfrac{BM . AU}{AB} = \dfrac{CM . CU}{AC} = CE$
10.07.2019 12:40
$$\frac{DM}{\sin \angle DBM}=\frac{BM}{\sin \angle ADM}=\frac{CM}{\sin \angle MEC}=\frac{EC}{\sin \angle EMC}=\frac{EC}{\sin 180^{\circ} - B} \implies DM=CE$$
01.09.2021 17:29
It's too easy.
01.09.2021 18:10
Note that $\measuredangle ECM=90^\circ+\measuredangle OAB=90^\circ+\measuredangle OMD=\measuredangle BMD$ and $\measuredangle MEC=\measuredangle MEA=\measuredangle MDB$, thus $\triangle ECM\sim\triangle DMB$, however as $MB=CM$, we get that $DM=EC$, we are done.
13.06.2022 20:59
Let $BC$ meet $AOM$ at $K$. Note that $\angle OAK + \angle OMK = \angle 180 \implies \angle OAK = \angle 90 \implies AK$ is tangent to $ABC$ so $\angle BMD = \angle KMD = \angle KAD = \angle ACB = \angle ACM$ and $\angle MEC = \angle MDB$ and $MC = MB$ so $MEC$ and $MDB$ are congruent so $CE = DM$.
29.06.2022 08:10
Let $A-$symmedian intersects $(ABC)$ at $K$. Notice that $(AOM)$ passes through the $K$ (inversion). Therefore $\triangle BDM\sim\triangle MEC$. $MB=MC\implies DM=CE$.