What fraction of the area of square with side of length $a$ does the shaded area represent?
I solved the problem of finding the fraction area of the triangle with sides of length $a$, $d$ and $e$; using the Pythagorean theorem, the sum of the angles of a triangle and a couple of trigonometric identities, but I'm wondering if there's a solution that doesn't resort to trigonometric identities.
The relation between $b$ and $a$ is simple: $b=\sqrt{2}a$.
Also,
$$\gamma=\frac{3\pi}{4} , \alpha+\beta+\gamma=\pi \Rightarrow \alpha+\beta=\pi/4$$
By the law of sines and the sine of a difference between angles:
$$\sin{\beta}=\sqrt{2}\sin{\alpha}=\sqrt{2}\sin{(\pi/4-\beta)}=\cos{\beta}-\sin{\beta}$$
So the relation between side lengths (and areas) follows:
$$\frac{\sin{\beta}}{\cos{\beta}}=\frac{1}{2}=\frac{e}{a} \Rightarrow \frac{ea/2}{a^2}=\frac{1}{4}$$
Thus, the area triangle of the triangle is one quarter of the small square, but I'm sure someone can come up with a more elegant solution.
Best Answer
Here is a simpler solution:
Use the fact that triangles △AOD and △BOC are congruent to obtain
$$BO=OD$$
Therefore, O is the midpoint of BD and the shaded area is $a^2/4$.