$\triangle ABC$ is a triangle and $D,E,F$ are points on $BC$, $CA$, $AB$ respectively. It is given that $BF=BD$, $CD=CE$ and $\angle BAC=48^{\circ}$. Find the angle $\angle EDF$ $ \textbf{(A) }64^{\circ}\qquad\textbf{(B) }66^{\circ}\qquad\textbf{(C) }68^{\circ}\qquad\textbf{(D) }70^{\circ}\qquad\textbf{(E) }72^{\circ} $
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wcao9311
04.06.2014 05:32
SMOJ wrote: $\triangle ABC$ is a triangle and $D,E,F$ are points on $BC$, $CA$, $AB$ respectively. It is given that $BF=BD$, $CD=CE$ and $\angle BAC=48^{\circ}$. Find the angle $\angle EDF$ $ \textbf{(A) }64^{\circ}\qquad\textbf{(B) }66^{\circ}\qquad\textbf{(C) }68^{\circ}\qquad\textbf{(D) }70^{\circ}\qquad\textbf{(E) }72^{\circ} $ Let $\angle BDF=x$, $\angle EDC=y$ We connect FD and ED to make 2 isosceles triangles. We use the facts of isosceles triangles, and supplementary angles to get that $540-2(x+y)+48=360$ then this simplifies to $x+y=118$ and $\angle EDF = 170-(x+y)=62$ QED.
TheMaskedMagician
04.06.2014 23:56
[asy][asy]
import olympiad;
size(5cm);
pair A,B,C,D,X, F;
B=origin; C=2*right; A=1.5*dir(70);
D=(B+C)/2;
X=(A+C)/2;
F=(A+B)/2;
draw(A--B--C--cycle);
dot(A);
dot(B);
dot(C);
dot(D);
dot(X);
dot(F);
draw(D--X);
draw(D--F);
label("$A$", A, N);
label("$B$", B, W);
label("$C$", C, E);
label("$D$", D, S);
label("$E$", X, E);
label("$F$", F, W);
label("$48^\circ$",shift(-0.1,-0.2)*A, E, fontsize(9pt));
label("$x$", shift(-0.1,-0.2)*X, E, fontsize(10pt));
label("$x$", shift(0.1,0)*D, N, fontsize(10pt));
label("$y$", shift(-0.01,0.2)*D, fontsize(10pt));
add(pathticks(C--D,1,0.5,0,5));
add(pathticks(B--D,1,0.5,0,5)); add(pathticks(B--D,1,0.6,7,5));
add(pathticks(C--X,1,0.5,0,5));
add(pathticks(B--F,1,0.5,0,5)); add(pathticks(B--F,1,0.6,7,5));
[/asy][/asy]
Let $\angle CDE=\angle CED=x$ and the desired $\angle EDF=y$. It follows that \[\angle BDF=\angle BFD=180-x-y\implies\angle AFD=x+y.\] Also, we have $\angle AED=180-x$. Since $AEDF$ is a quadrilateral, the angles sum to $360^\circ$. So we have \begin{align*}48+(180-x)+(x+y)+y=360 \\ &\implies 228+2y=360 \\ &\implies 114+y=180 \\ &\implies y=180-114=\boxed{\textbf{(B) }66^\circ.}\end{align*}
wcao9311
05.06.2014 01:29
@themaskedmagician thanks. I think I did $(588-360)/2$ wrong and got 118 instead of 114.