$ABCD$ is a square. $E$ and $F$ are points respectively on $BC$ and $CD$ such that $\angle EAF = 45^\circ$. $AE$ and $AF$ cut the diagonal $BD$ at $P,Q$ respectively. Find $\frac{[\Delta AEF]}{[\Delta APQ]}$ .
What I Tried: Here is a picture :-
I tried to use a bit of angle-chasing, and could only conclude that $\Delta BPE \sim \Delta QDF$, but this does not seem to be useful anywhere.
Then I used Geogebra to look for some results. There I found that $\angle APQ = \angle AFE$, concluding that $\Delta APQ \sim \Delta AFE$ .
Can anybody explain which is that so?
I put $\angle AFE = (90 – x)^\circ$ and $\angle AEF = (45 + x)^\circ$, to my surprise, the results worked, but why?
Also the answer to the problem is $2$, but how do you find the result?
Can anyone help?
Best Answer
Since $\angle QAP = \angle EBP$, $ABEQ$ is cyclic quadrilateral.
Rotate $\triangle ADF$ about $A$ by $90°$ clockwise. Let F' be the new point of F.
Since $\triangle ADF \cong \triangle ABF'$, $\angle AQB = \angle AEB = \angle AEF$.
Therefore, $\triangle APQ \sim \triangle AFE$. It implies $AP×AE = AQ×AF$.
Join $EQ$. $\angle AQE = 180° - 45° - \angle AEQ = 135° - \angle ABQ = 90°$.
Then, $AQ = EQ$ implies $\frac{AQ}{AE} = \frac{1}{\sqrt{2}}$.
Similarly, $\frac{AP}{AF} = \frac{1}{\sqrt{2}}$.
Finally, $\frac{[\triangle AFE]}{[\triangle APQ]} = \frac{0.5}{0.5}\frac{\sqrt{2}}{1}\frac{\sqrt{2}}{\sqrt{1}}\frac{sin\angle FAE}{sin\angle PAQ} = 2$