Point $D$ lies in $\triangle ABC$ and $BD=CD$,$\angle BDC=120$. Point $E$ lies outside $ABC$ and $AE=CE,\angle AEC=60$. Points $B$ and $E$ lies on different sides of $AC$. $F$ is midpoint $BE$. Prove, that $\angle AFD=90$
Problem
Source: Moscow Math Olympiad 2017, Grade 11, P10
Tags: geometry
aryanna30
25.04.2017 17:47
My solution (with calculation) : Let $BC=a , AC=b , \angle BCA=C $ so ${AB}^2=a^2+b^2-2ab cos \angle C , DC=\tfrac{a}{\sqrt 3} ,{BE}^2= a^2+b^2-2ab cos \angle (C+60) , {DE}^2= \tfrac{a^2}{3} + b^2 - \tfrac{2 \sqrt 3 ab cos\angle (C+30)}{3} , {AD}^2=\tfrac{a^2}{3} + b^2 - \tfrac{2 \sqrt 3 ab cos\angle (C-30)}{3}$ so we need to prove that ${DB}^2+{DE}^2+c^2+b^2=2{AD}^2$ $\rightarrow$ $\tfrac{2a^2}{3} + b^2 - \tfrac{2 \sqrt 3 ab cos\angle (C+30)}{3}-a^2-b^2+2ab cos \angle (C+60)+a^2+2b^2-2ab cos \angle C=\tfrac{2a^2}{3} +2b^2-2ab(\tfrac{2 \sqrt 3 cos\angle (C+30)}{3} -cos \angle(C+60) + cos\angle C)$ so we need to prove that $\tfrac{2 \sqrt 3 cos\angle (C+30)}{3} -cos \angle(C+60) + cos\angle C=\tfrac{4 \sqrt 3 cos\angle (C-30)}{3}$ assume that $cos \angle C=x , sin \angle C=y$ so $\tfrac{3x}{2}-\tfrac{\sqrt 3 y}{6}=x+\tfrac{x}{2}-\tfrac{\sqrt3 y}{2}+\tfrac{\sqrt3 y}{3}$ so we are done!
aryanna30
25.04.2017 18:07
Simple solution: Let $A'$ be the reflection of $A$ in point $F$. $BA'$ cuts $AC$ at $G$. so $AEA'B$ be the parallelogram.$BA' \parallel AE$ so $\angle BGC=\angle BDC=120$ so $B,D,G,C$ are concyclic. $\angle DBG=\angle DCA$,$AC=AE=BA' , BD=CD$ so $\triangle BDA' \cong CDA$ so $DA'=DA$ so done.