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

(N/A) Let $\triangle ABC$ be a right-angled triangle with $\angle A = 90^\circ$,$AC = y$,and $AB = x$.
Three equilateral triangles $\triangle AEC$,$\triangle AFB$,and $\triangle CBD$ are drawn on sides $AC$,$AB$,and $BC$ respectively.
Let the areas of these equilateral triangles be $A_1$,$A_2$,and $A_3$ respectively.
We need to prove that $A_3 = A_1 + A_2$.
In $\triangle ABC$,by Pythagoras theorem: $BC^2 = AC^2 + AB^2 = y^2 + x^2$.
The area of an equilateral triangle with side $s$ is given by $\frac{\sqrt{3}}{4} s^2$.
$A_1 = \text{Area}(\triangle AEC) = \frac{\sqrt{3}}{4} AC^2 = \frac{\sqrt{3}}{4} y^2$ ... $(i)$
$A_2 = \text{Area}(\triangle AFB) = \frac{\sqrt{3}}{4} AB^2 = \frac{\sqrt{3}}{4} x^2$ ... $(ii)$
$A_3 = \text{Area}(\triangle CBD) = \frac{\sqrt{3}}{4} BC^2 = \frac{\sqrt{3}}{4} (y^2 + x^2)$ ... $(iii)$
From $(i)$,$(ii)$,and $(iii)$:
$A_3 = \frac{\sqrt{3}}{4} y^2 + \frac{\sqrt{3}}{4} x^2 = A_1 + A_2$.
Hence,the area of the equilateral triangle on the hypotenuse is equal to the sum of the areas of the equilateral triangles on the other two sides.