Prove that the area of the semicircle drawn on the hypotenuse of a right-angled triangle is equal to the sum of the areas of the semicircles drawn on the other two sides of the triangle.

(N/A) Let $ABC$ be a right triangle,right-angled at $B$ with $AB = y$ and $BC = x$.
Three semicircles are drawn on the sides $AB$,$BC$,and $AC$ respectively,with diameters $AB$,$BC$,and $AC$.
Let the areas of the semicircles with diameters $AB$,$BC$,and $AC$ be $A_1$,$A_2$,and $A_3$ respectively.
To prove: $A_3 = A_1 + A_2$.
Proof: In $\triangle ABC$,by the Pythagoras theorem,
$AC^2 = AB^2 + BC^2$
$AC^2 = y^2 + x^2$
$AC = \sqrt{y^2 + x^2}$
We know that the area of a semicircle with diameter $d$ is $\frac{\pi}{2} \left(\frac{d}{2}\right)^2 = \frac{\pi d^2}{8}$.
Therefore,the area of the semicircle drawn on $AC$ is:
$A_3 = \frac{\pi (AC)^2}{8} = \frac{\pi (y^2 + x^2)}{8}$ .....$(i)$
Now,the area of the semicircle drawn on $AB$ is:
$A_1 = \frac{\pi (AB)^2}{8} = \frac{\pi y^2}{8}$ .....$(ii)$
And the area of the semicircle drawn on $BC$ is:
$A_2 = \frac{\pi (BC)^2}{8} = \frac{\pi x^2}{8}$ .....$(iii)$
Adding equations $(ii)$ and $(iii)$:
$A_1 + A_2 = \frac{\pi y^2}{8} + \frac{\pi x^2}{8} = \frac{\pi (y^2 + x^2)}{8}$
From equation $(i)$,we see that $A_1 + A_2 = A_3$.
Hence,the area of the semicircle on the hypotenuse is equal to the sum of the areas of the semicircles on the other two sides.