Let $ABC$ be a triangle, where $AB = AC$ and $BC = 12$. Let $D$ be the midpoint of $BC$. Let $E$ be a point in $AC$ such that $DE \perp AC$. Let $F$ be a point in $AB$ such that $EF \parallel BC$. If $EC = 4$, determine the length of $EF$.
Problem
Source:
Tags: Asymptote
TheMaskedMagician
01.09.2014 00:41
[asy][asy] import cse5; import olympiad; pair B=(0,0), C=(12,0), A=(6,10), D=(B+C)/2, EE=foot(D,A,C), F=foot(D,A,B),X=(EE+F)/2; D(MP("A",A,N)--MP("B",B,W)--MP("C",C,E)--cycle); D(MP("D",D,S)--MP("E",EE,E)--MP("F",F,W)--cycle); D(A--MP("X",X,S)); D(rightanglemark(D,F,B,20)); D(rightanglemark(D,EE,C,20)); D(rightanglemark(A,X,EE,20)); label("$6$",(B+D)/2,S); label("$6$",(C+D)/2,S); label("$4$",(EE+C)/2,E); label("$4$",(B+F)/2,W); label("$x$",(A+EE)/2,E); label("$x$",(A+F)/2,W); label("$\frac{y}{2}$",(EE+X)/2,N); [/asy][/asy] Since $AB=AC$, it follows that $BF=4$ as well, and that $AE=AF$. Let $AE=AF=x$, $X$ be the foot of the perpendicular from $A$ to $EF$, and $y$ be the length of $EF$. Since $EF\parallel BC$, then $\triangle AFE\sim\triangle ABC$. Then, \[\frac{x}{x+4}=\frac{y}{12}.\] The similarity also tells us that $\angle AEF=\angle DCE$. Thus, $\triangle AXE\sim\triangle DEC$ by AA similarity. So \[\frac{x}{6}=\frac{y}{8}.\] From our two proportions, we have \begin{align}12x&=xy+4y\\ 8x&=6y.\end{align} Then $x=\frac{3y}{4}$ and then \[9y=\frac{3y^2}{4}+4y\implies y=\frac{20}{3}.\] Thus, $EF=\boxed{\frac{20}{3}}$. EDIT: Oops didnt see Sharky's hint but I had the same idea.