Question

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

Answer 1

Let $△APB$ and parallelogram $ABCD$ lie on base $AB$ and between parallels $AB$ and $PC$.
To show area $△APB=\frac{1}{2}$ Area $\left(ABCD\right)$
Now, draw BQ || AP.
Then $ABQP$ is a parallelogram.
Now area $ABQP=$ Area $ABCD$.....(They are on same base $AB$ and between same parallels $AB$ and $PC)$

$$\begin{array}{cc}& \Rightarrow △PAB\cong △BQP\\ & \text{Area}PAB=\text{Area}BQP\\ & =\frac{1}{2}\text{Area}ABQP\\ & =\frac{1}{2}\text{Area}ABCD\end{array}$$