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Answer :

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.

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