Question

In parallelogram ABCD, p is a point on side AB and Q is a point on side BC.
Prove that Area ($\Delta$ AQD) = Area ($\Delta$ APD) + Area ($\Delta$ CPB)

Answer 1

$Ar\left(△APD\right)=Ar\left(△APC\right)$ $(\because$ Area of triangle with same base $AP$ and between same set of parallel lines are equal $)$

$Ar\left(△APD\right)=Ar\left(△CPB\right)=Ar\left(△APC\right)=Ar\left(△CPB\right)=Ar\left(△ABC\right)=\frac{1}{2}$ Area of // gm $ABCD$

$(\because AC$ is diagnol and diagnols divides it into equal areas $)$

$Ar\left(△AQD\right)=Ar\left(△APD\right)+Ar\left(△CPB\right)$