Question

If a triangle and a parallelogram are on the same base and between the same parallels, then prove that area of triangle is half the area of parallelogram.

Answer 1

Draw $AL\perp BC$ and $DM\perp BC$, meeting BC produced in M.

Since, A, E and D are collinear and $BC\parallel AD$

Therefore,

$AL=DM$

Now,

$ar\left(△ABC\right)=\frac{1}{2}(BC\times AL)$

$ar\left(△ABC\right)=\frac{1}{2}(BC\times DM)$

$ar\left(△ABC\right)=\frac{1}{2}ar\left({\parallel }^{gm}BCDE\right)$