In my isosceles triangle $\vartriangle ABC$ with $AB = CA$, we draw $D$ the midpoint of $BC$. Let $E$ be a point on $AC$ such that $\angle CDE = 60^o$ and $M$ the midpoint of $DE$. Prove that $\angle AME = \angle BMD$.
Problem
Source: 2022 Mathematics Regional Olympiad of Mexico West P3
Tags: equal angles, isosceles, geometry
Tsikaloudakis
08.10.2022 20:57
solution later
sunken rock
15.10.2022 11:35
Maybe the following lemma will be easier... Quote: Let $D$ be midpoint of the base $BC$ of the isosceles triangle $ABC$, $E$ that point onto the segment $AC$ so that $\widehat{CDE}=60^\circ, M$ midpoint of $DE$, while $F,N$ are reflections of $E,M$ respectively about $AD$ and $\{\ P\ \}\in BM\cap AN$. Prove that $m(\widehat{APM})=60^\circ$. Best regards, sunken rock
sunken rock
16.10.2022 09:30
Tsikaloudakis wrote: Except for possible error: How did you get $AK,BM,EL$ concurrent? Best regards, sunken rock
Tsikaloudakis
16.10.2022 16:55
I will prove this in a future post.
sunken rock
27.10.2022 23:00
sunken rock wrote: Maybe the following lemma will be easier... Quote: Let $D$ be midpoint of the base $BC$ of the isosceles triangle $ABC$, $E$ that point onto the segment $AC$ so that $\widehat{CDE}=60^\circ, M$ midpoint of $DE$, while $F,N$ are reflections of $E,M$ respectively about $AD$ and $\{\ P\ \}\in BM\cap AN$. Prove that $m(\widehat{APM})=60^\circ$. Best regards, sunken rock
$P$ is Fermat point for triangles $BDN$ and $ABD$
Best regards, sunken rock
sunken rock
29.10.2022 17:10
A solution at https://www.facebook.com/photo.php?fbid=8224005504339606&set=p.8224005504339606&type=3