AMC 12 Mock Exam 4 Problem 17

Let $\alpha=\angle BAE$, so $\tan\alpha=\frac{BE}{AB}=\frac13$ and $\angle DAF=90^\circ-45^\circ-\alpha=45^\circ-\alpha$. Then $$DF=\tan(45^\circ-\alpha)=\frac{1-\frac13}{1+\frac13}=\frac12.$$ Rotate $\triangle ADF$ by $90^\circ$ about $A$ so that $D$ goes to $B$; $F$ goes to a point $F'$ on line $BC$ on the opposite side of $B$ from $C$, with $BF'=DF=\frac12$ and $AF'=AF$. Then $\angle EAF'=\angle EAB+\angle BAF'=\alpha+(45^\circ-\alpha)=45^\circ=\angle EAF$, so $\triangle AEF\cong\triangle AEF'$ (SAS). Triangle $AEF'$ has base $EF'=BE+BF'=\frac13+\frac12=\frac56$ on line $BC$ and height $AB=1$, so $$[AEF]=[AEF']=\frac12\cdot\frac56\cdot1=\boxed{\tfrac5{12}}.$$ (Check with coordinates $A=(0,0)$, $E=(1,\tfrac13)$, $F=(\tfrac12,1)$: the area is $\frac12\left|1\cdot1-\tfrac13\cdot\tfrac12\right|=\frac5{12}$.)