The ratio of the area of the equilateral triangle described on the side of a square is half the area of the equilateral triangle described on its diagonal

Given : A square ABCD. Equilateral triangle $\mathrm{△}\mathrm{BCE}\text{ and }\mathrm{△}\mathrm{ACF}$have been described on side BC and diagonal AC respectively.

Proof : Since $\mathrm{△}\mathrm{BCE}\text{ and }\mathrm{△}\mathrm{ACF}$ are equilateral. Therefore, they are equiangular (each angle being equal to 60°) and hence
$\begin{array}{l}\mathrm{\Delta BCE}~\mathrm{\Delta ACF}\\ \Rightarrow \frac{\mathrm{Area}\left(\mathrm{\Delta BCE}\right)}{\mathrm{Area}\left(\mathrm{\Delta ACF}\right)}=\frac{{\mathrm{BC}}^2}{{\mathrm{AC}}^2}\\ \Rightarrow \frac{\mathrm{Area}\left(\mathrm{\Delta BCE}\right)}{\text{ Area }\left(\mathrm{\Delta ACF}\right)}=\frac{{\mathrm{BC}}^2}{(\sqrt{2}\mathrm{BC}{)}^2}\left[\begin{array}{c}\because \mathrm{ABCD}\text{ is a square }\\ \therefore \text{ diagonal }=\sqrt{2}\left(\text{ si de }\right)\Rightarrow \mathrm{AC}=\sqrt{2}\mathrm{BC}]\end{array}\right.\\ \Rightarrow \frac{\text{ Area }\left(\mathrm{\Delta BCE}\right)}{\text{ Area }\left(\mathrm{\Delta ACF}\right)}=\frac{1}{2}\end{array}$