Let $ ABC$ be a triangle, $ D$ be the midpoint of $ BC$, $ E$ be a point on segment $ AC$ such that $ BE=2AD$ and $ F$ is the intersection point of $ AD$ with $ BE$. If $ \angle DAC=60^{\circ}$, find the measure of the angle $ FEA$.
Problem
Source: Central American Olympiad 2002, problem 4
Tags: geometry, parallelogram, trapezoid
Brut3Forc3
30.12.2009 10:37
Let $ G$ be the midpoint of $ BE$. $ DG$ is a midline of triangle $ BEC$, so $ \overline{DG} \parallel \overline{AC}$, so $ \triangle FGD \sim \triangle FEA$. Also, $ GE=\frac{1}{2}BE=AD$, which implies that $ AF=FE$. Thus, $ \angle AEF= \angle EAF = 60^\circ$.
dgreenb801
30.12.2009 17:29
Let $ G$ be the point such that $ ABGC$ is a parallelogram, so $ AG = 2AD = BE$ and $ \angle AGB = 60$. Let the parallel through $ A$ to $ GC$ meet $ BG$ at $ H$. Then $ AHBE$ is a parallelogram, so $ AG = BE = AH$, so since $ \angle AGH = 60$, $ \triangle AGH$ is equilateral. Thus, $ \angle AHB = \angle AEF = 60$.
sunken rock
29.02.2012 20:58
@dgreenb801: Obviously, $ABGE$ is an isosceles trapezoid ($BG\parallel AE, AG=BE$), done. Best regards, sunken rock
BttfUS
01.03.2012 00:41
Let$X$ be a point on the ray$AD$,such that:$XD=AD$ So$AC//PX$ $AX=BE$ So$ABEX$ is a isosceles trapezoid So$\angle FEA=\angle FAE=\frac{\pi}{3}$
Diehard
01.03.2012 01:20
Sorry if this is any of the above posters' solutions: Let $X$ be the midpoint of $BE$. Then, since $AD=XE$ and $XD$ is parallel to $AC$, $AEDX$ must be an isosceles trapezoid and $\triangle AEF$ is isosceles so $\angle FEA=60$.
hatchguy
08.03.2012 04:24
I think the most natural solution hasn't been posted. Let $G$ be the midpoint of $EC$. Then $DG = \frac{BE}{2} = AD$ and since $\angle DAG = \angle DAC = 60$ we have that $ADG$ is equilateral and therefore $\angle DGA = 60$. Since $FE \parallel DG$ we have $\angle FEA = \angle DGA = 60$ and we are done.