Question

In the figure, $ABCD$ is a quadrilateral. $A$ line through $D$, parallel to $AC$, meets $BC$ produced in $P$. prove that area $\left(△ABP\right)=area\left(quadrilateralABCD\right)$.

• A. True
• B. False

Answer 1

The correct option is A True

Solution :-

Since in $△APC$ and $△ADC$ are lie on the

same base AC and between the same

parallel line AD and CP.

So, $ar\left(APC\right)=ar\left(ADC\right)$

By adding both sides $△ABC$ ... (1)

so we get $ar\left(APC\right)+ar\left(ABC\right)=ar\left(ADC\right)+ar\left(ABC\right)$

$\Rightarrow ar\left(ABP\right)=ar\left(quad\right(ABCD\left)\right)$

H.P