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!
Given an isosceles right-triangle $\triangle ABC$ with 3 squares on the hypotenuse, find the area of the largest square.
contest-matheuclidean-geometrygeometrytrianglestrigonometry
Related Solutions
This is my own approach. I'll add a brief explanation too:
This is the procedure I followed:
1.) Label the triangle $ABC$ and mark all the appropriate angles and side lengths. Notice that since $\triangle ABC$ is obtuse, its circumcenter must lie outside of the triangle itself. Locate circumcenter of $\triangle ABC$ outside and call it point $E$. Connect all the vertices of $\triangle ABC$ to $E$ via $AE$, $BE$ and $CE$. Notice that this means $AB=BE=AE=CE=CD$ because $\triangle ABE$ is equilateral.
(for further explanation of above, point $E$ is a circumcenter of $\triangle ABC$, in that case, $\angle AEB$ must be twice the measure of $\angle ACB$ (inscribed angle theorem), which means it'll be $60$. Because $\angle AEB=60$ and know that $AE=BE$, it follows that $\triangle ABE$ is equilateral)
2.) Notice also that $\angle EAC=\angle ECA=6$. Connect point $D$ with $E$ via segment $DE$. Notice that because segment $CD=CE$ and $\angle ECD=36$, this implies that $\angle EDC=\angle DEC=72$. However, notice that $\angle EBD=96-60=36$, therefore, $\angle BED$ is also $36$. But this implies that segment $BD=DE$
3.) Lastly, notice that this means $\triangle ABD$ is congruent to $\triangle AED$ via the SAS property. This means $\angle BAD=\angle EAD=x$. This means that $2x=60$, therefore, $x=30$.
A very simple approach, based on your picture, using the law of sines:
$$\angle CAO=40-x,\angle COA=140^{\circ} \implies \frac{\sin (40-x)^{\circ}}{\sin 140^{\circ}}=\frac {\sin (20+x)^{\circ}}{\sin 40^{\circ}}=\frac{\sin (20+x)^{\circ}}{\sin 140^{\circ}}$$
$$\implies \sin (40-x)^{\circ}=\sin (20+x)^{\circ}\implies 40-x=20+x\implies x=10^{\circ}.$$
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Best Answer
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,
$$ CD^2+BE^2=DE^2 $$