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In square $ABCD$ point $E$ lies on $BC$ and point $F$ lies on $DE$ such that: $DF=BF , EF=EB$.Calculate the measure of angle $DFA$.

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It's easily seen that $\angle BFE=2\angle BDF$.But I can't go on...

Hamid Reza Ebrahimi
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2 Answers2

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Notice that $\angle BFE + \angle BDF = \frac {\pi}{4}$, together with $\angle BFE=2\angle BDF$ makes a system of equations, easy to solve it.

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It is $75^{\circ}$.

If $\angle FBE =x$ then $\angle BFE=x$ and $\angle DEC = 2x$. Thus $\angle DBF =45-x$ and $\angle CDE =90-x$. Now $\angle FDB =45-x$. Thus $90-2x + 45-x =45$so $x=30$ thus $\angle DFA =75$ since AF is angle bisector.

nonuser
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