Question

If p and q are perpendiculars from the angular points A and B of $△ABC$ drawn to any line through the vertex C, then prove that $a^2{b}^2{\sin }^{2}C=a^2{p}^2+b^2{q}^2-2abpq\cos C$

Answer 1

Let $\Delta$ be the area of $△ABC$
Then $\Delta =\frac{1}{2}ap=\frac{1}{2}bq$
$\Rightarrow p=\frac{2\Delta }{a};q=\frac{2\Delta }{b}$
Consider $a^2{p}^2+b^2{q}^2-2abpq\cos C$
$=a^2\left(\frac{4{\Delta }^2}{a^2}\right)+b^2\left(\frac{4{\Delta }^2}{b^2}\right)-2ab\frac{4{\Delta }^2}{ab}\cos C$
$=8{\Delta }^2-8{\Delta }^2\cos C$
$=16{\Delta }^2{\sin }^2\frac{C}{2}$