Given an isosceles right-triangle $\triangle ABC$ with 3 squares on the hypotenuse, find the area of the largest square. As the title suggests, the question is to solve for the area of the largest square in the following figure given the area of 2 smaller ones. This is a pretty fun problem and I want to see if there are any more ways to solve it, such as with trigonometry. As always, I'll post my own approach as an answer below!

 A: The large square's area is always the sum of the other two squares' areas.

Say $\triangle ABC$ is our isosceles right triangle, with right angle at $A$. $D$ and $E$ are on hypotenuse $BC$, and $\angle DAE$ measures $45^\circ$. Reflect $C$ across line $AD$ to find point $F$ — that is, $\angle CAD = \angle DAF$ and $AC=AF$. Then
$$ 90^\circ = \angle CAB = \angle CAD + \angle DAE + \angle EAB = \angle CAD + \angle EAB + 45^\circ $$
$$ 45^\circ = \angle CAD + \angle EAB $$
$$ 45^\circ = \angle DAE = \angle DAF + \angle FAE $$
Since by construction $\angle CAD = \angle DAF$, this gives $\angle EAB = \angle FAE$. We then have (SAS) two congruent triangle pairs, $\triangle ACD \cong \triangle AFD$ and $\triangle ABE \cong \triangle AFE$.
The congruences give $\angle DFA = \angle DCA = 45^\circ$ and $\angle AFE = \angle ABE = 45^\circ$, so $\angle DFE = \angle DFA + \angle AFE = 90^\circ$, and $\triangle DEF$ is a right triangle. By the Pythagorean theorem, $DF^2+EF^2=DE^2$. By the congruent triangles again, $CD=DF$ and $BE=EF$. Finally,
$$ CD^2+BE^2=DE^2 $$
A: 
Here is an alternative approach that leads to the same conclusion as that of aschepler.
Rotate $\Delta BAE$ $  90^o$ about $B$ to obtain $\Delta BCF$ as shown.
Note that
(1) $AE=CF$
(2) $\Delta BDE \cong \Delta BDF $ (SAS)$\implies DE=DF$
(3) $\angle DCF=90^o$
From Pythagoras Theorem,
$CF^2+CD^2=DF^2$
Hence  $AE^2+CD^2=DE^2$
A: According to the Goku's figure, let $AB=BC=m$, $AE=a$, $EF=z$, $FC=b$. By Law of sines, in $\triangle ABE$ we have $\frac{a}{\sin\alpha}=\frac{m}{\sin(135-\alpha)}$ and in $\triangle BFC$ we have $\frac{b}{\sin(45-\alpha)}=\frac{m}{\sin(90+\alpha)}$. By eliminating $m$ we get $\tan(2\alpha)=\frac{a}{b}.$
Here, $a=2$, $b=2\sqrt{3}$, so $\alpha=15$.
By law of sines in $\triangle BFC$, $\frac{b}{\sin30}=\frac{BF}{\sin45}$ and thus $BF=2\sqrt{6}$.
By law of sines in $\triangle ABF$, we have $\frac{BF}{\sin45}=\frac{z+a}{\sin60}$ and $z+2=6$ and thus $z=EF=4$. So the area of the big square is $4^2=16$.
A: This is my method of solving it. I'll explain below:

Here's my explanation:
1.) Label the triangle as $\triangle ABC$ with two points $E$ and $F$ on the hypotenuse. Notice that since $\angle FBE=45$, we can set $\angle FBC=\alpha$ and that means that $\angle EBA=45-\alpha$. In other words $\angle FBC+\angle EBA=45$. Using this fact, we can rotate $\triangle BFC$ BG $90$ counterclockwise to form a new triangle $\triangle BDA$ outside of $\triangle ABC$ which will be congruent to $\triangle BFC$
2.) This means that $BD=BF$ and that $\angle DBA=45-\alpha$. However, this implies that $\angle DBE=45$, therefore, we can say that $\triangle DBE$ is congruent to $\triangle EBF$. By joining $D$ and $E$ via $DE$, we can say that $DE=EF$ as well. Now, since the areas of the given squares are $4$ and $12$ respectively, their side lengths would be $2$ and $2\sqrt{3}$ respectively, as well. Which means that $DA=2\sqrt{3}$ and $AE=2$. Notice that since $\angle DAB=45$ and $\angle BAC=45$, we know that $\angle DAE=90$. This means that $\triangle DAE$ is a right-angle triangle with sides $2$ and $2\sqrt{3}$, and thus, by applying the Pythagorean theorem we have:
$$DE^2=(2)^2+(2\sqrt{3})^2=16$$
This means that $DE^2=EF^2=16$, which gives us the answer that the area of the largest square is $16$.
