It can be observed that \\( \ riangle B Q C \\) and paralleogram ABCD lie on the same base BC and these are between the same parallel lines AD and BC.
\\( \ herefore \ ext { Area }(\\Delta B Q C)=\\frac{1}{2} \ ext { Area (ABCD) } \\dots \\) (1)
Similarly, \\( \ riangle \\mathrm{APB} \\) and parallelogram ABCD lie on the same base AB and between the same parallel lines AB and DC.
\\( \ herefore \ ext { Area }(\\Delta \\mathrm{APB})=\\frac{1}{2} \ ext { Area (ABCD) } \\ldots \\) (2)
From equation (1) and (2), we obtain
Area \\( (\\Delta B Q C)=\ ext { Area }(\\Delta A P B) \\)