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# 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.

Given, ABCD is a square whose one diagonal is AC. $$\triangle$$ APC and $$\triangle$$ BQC are two equilateral triangles described on the diagonals AC and side BC of the square ABCD.

Area($$\triangle$$ BQC) = $${{1} \over {2}}$$ Area($$\triangle$$ APC)

Since, $$\triangle$$ APC and $$\triangle$$ BQC are both equilateral triangles, as per given,
$$\triangle$$ APC ~ $$\triangle$$ BQC [AAA similarity criterion]
area($$\triangle$$ APC)/area($$\triangle$$ BQC) = $$\frac{AC^2}{BC^2} = \frac{AC^2}{BC^2}$$

Since, Diagonal = $$\sqrt{2}$$ side = $$\sqrt{2}$$ BC
$$({{\sqrt{2} BC } \over {BC}})^2 = 2$$
area($$\triangle$$ APC) = 2 × area($$\triangle$$ BQC)
area($$\triangle$$ BQC) = $${{1} \over {2}}$$area($$\triangle$$ APC)
Hence, proved.