Question

Prove that area of an equilateral triangle formed any side of a square is half the area of an equilateral triangle formed at the diagonal of same square.

Answer 1

$ABCD$ is a square whose our side is $AB$ and diagonal is $AC$. An equilateral triangles $ABE$ and $ACF$ are formed on the sides of$AB$ and $AC$
Proof: In right angle $△ABC$
${AC}^2={AB}^2+{BC}^2$ (by Pythagoras theorem)
${AC}^2={AB}^2+{AB}^2$
${AC}^2=2{AB}^2$
$AC=\sqrt{2}AB$
Area of equilateral triangle $ABE$ formed on side $AB=\frac{\left({AB}^2\right)\sqrt{3}}{4}$
area of equilateral triangle $ACF$ formed by hypotenuse $AC=\frac{{\left(AC\right)}^2\sqrt{3}}{4}$
$\frac{ar.\left(△ABE\right)}{ar.\left(△ACF\right)}=\frac{\frac{{\left(AB\right)}^2\sqrt{3}}{4}}{\frac{{\left(AC\right)}^2\sqrt{3}}{4}}=\frac{{\left(AB\right)}^2\sqrt{3}}{{\left(AC\right)}^2\sqrt{3}}=\frac{{\left(AB\right)}^2}{{\left(AC\right)}^2}={\left(\frac{AB}{AC}\right)}^2={\left(\frac{AB}{\sqrt{2}AB}\right)}^2={\left(\frac{1}{\sqrt{2}}\right)}^2=\frac{1}{2}$

$ar.△ABE=\frac{1}{2}ar.△ACF$

Proved