Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

(N/A) Let $ABCD$ be a square of side $a$.
Therefore,its diagonal $= \sqrt{2} a$.
Two equilateral triangles are formed: $\Delta ABE$ (on side $AB$) and $\Delta DBF$ (on diagonal $DB$).
Side of equilateral triangle $\Delta ABE$ described on side $AB = a$.
Side of equilateral triangle $\Delta DBF$ described on diagonal $DB = \sqrt{2} a$.
Since all equilateral triangles are similar,the ratio of their areas is equal to the square of the ratio of their corresponding sides.
$\frac{\text{Area of } \Delta ABE}{\text{Area of } \Delta DBF} = \left( \frac{a}{\sqrt{2} a} \right)^2 = \left( \frac{1}{\sqrt{2}} \right)^2 = \frac{1}{2}$.
Thus,$\text{Area of } \Delta ABE = \frac{1}{2} \times \text{Area of } \Delta DBF$.