Interesting Problem

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If $ABCD$ has side $x$, then $BX \,=\, x\cos\theta$ and $DY = x\sin\theta$. Since $\angle ABX = 180^\circ-\theta$ and $\angle ADY = 90^\circ+\theta$, we deduce from the Cosine Rule that $\begin{array}{rcl} y^2 & = & x^2 + x^2\cos^2\theta - 2x^2\cos\theta\cos(180^\circ-\theta) \; = \; x^2(1 + 3\cos^2\theta) \\ z^2 & = & x^2 + x^2\sin^2\theta - 2x^2\sin\theta\cos(90^\circ+\theta) \; = \; x^2(1 + 3\sin^2\theta) \end{array}$ Using the Sine Rule on $ABX$ and $ADY$, $\begin{array}{rcl} \sin\alpha & = & \frac{x\cos\theta \times \sin(180^\circ-\theta)}{y} \; = \; \frac{\sin\theta\cos\theta}{\sqrt{1+3\cos^2\theta}} \\ \sin\beta & = & \frac{x\sin\theta \times \sin(90^\circ+\theta)}{z} \; = \; \frac{\sin\theta\cos\theta}{\sqrt{1+3\sin^2\theta}} \end{array}$ and hence $\cos\alpha \; = \; \frac{1 + \cos^2\theta}{\sqrt{1 + 3\cos^2\theta}} \qquad \cos\beta \; = \; \frac{1 + \sin^2\theta}{\sqrt{1 + 3\sin^2\theta}}$ Hence $\cos(\alpha+\beta) \; = \; \frac{(1+\cos^2\theta)(1+\sin^2\theta) - \sin^2\theta\cos^2\theta}{\sqrt{(1 + 3\cos^2\theta)(1 + 3\sin^2\theta)}} \; = \; \frac{2}{\sqrt{(1 + 3\cos^2\theta)(1 + 3\sin^2\theta)}}$ Now $\gamma = 90^\circ - \alpha-\beta$ and so $\sin\gamma = \cos(\alpha+\beta)$. Thus $|AEF| \; = \; \tfrac12yz\sin\gamma \; = \; \frac{yz}{\sqrt{(1 + 3\cos^2\theta)(1 + 3\sin^2\theta)}} \; = \; x^2 \; = \; |ABCD|$

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There's no need to bust out the complicated trigonometry. ;)

We note that $X'A=XA$, $AB=BC=CD=DA$ and $Y'A=YA$. Also $|BCV|=|XBA|=|ADY|$, which can be easily proved by trigonometry--noting that $|BCV| = \frac{1}{2} \times BV \times BC = \frac{1}{2} \times BX \times BA =|XBA|$ and the same can be done for triangle $ADY$.

We construct $X'B'A$ by rotating $XBA$ $90$ degrees clockwise about $A$ and $X_1'B'C$ by rotating $XBA$ $90$ degrees clockwise about $B$. Similarly we construct $Y'D'A$ by rotating $YDA$ $90$ degrees anticlockwise about $A$ and $Y_1'B'C$ by rotating $YDC$ $90$ degrees anticlockwise about $D$.

Rotation preserves area and congruency, and we see that by angle chasing, $X'B'$ is equal in length and parallel to $X_1'B$ and the same with the pairs $B'D'$ and $BD$, $Y'D'$ and $Y_1'D$. Therefore we see that $X'Y'D'AB'$ is congruent to $X_1'Y_1'DCB$.

More importantly we also see that $|BDHV|=|X'Y'D'B'|=|X_1'Y_1'DB|$ as all three share the same height $VH$ and lengths $BV$ and $DH$.

Therefore

$|X'Y'A|=|X'Y'D'AB'|-|X'B'A|-|Y'D'A|=|X_1'Y_1'DCB|-|X_1'BC|-|CDY_1'|$

$=|X_1'Y_1'DB|+|BCD|-|BVC|-|DCH|=|BDHV|+|BCD|-|BVC|-|DCH|$

$=|BCD|+|BVC|+|DCH|+|BCD|-|BVC|-|DCH|=2|BCD|=|ABCD|$

And we are done! ;)

This is obvious, but the area of ABCD equals the sum of the areas of the two smallest squares. I don't know if that would help though.

EDIT: I played with the problem a little bit. Label the side of the smallest square to the side of the biggest square a-e, respectively. After tinkering with the Pythagorean Theorem, I have come to the conclusion that $5c^2=d^2+e^2$, so if you can prove that $\dfrac{d^2+e^2}{5}$ is the area of $\Delta AEF$ then you are done.

Do the two lower squares have coincident bottom sides, and does C lie on the extension of both bottom lines?

wow so complex

teach me your ways master