Question

P is any point on the diagonal BD of the parallelogram ABCD. Prove that $\Delta APD=\Delta CPD$ in area.

Answer 1

R.E.F image

Given: Parallelogram $ABCD$

$P$ any point on $BD$

To prove $ar\left(△APD\right)=ar\left(△CPD\right)$

Construction : Draw $O$ mid-pt of $BD$

Now, $AC$ & $BD$ intersect at $O$.

Since daigonal of parallelogram bisect each other

$O$ is mid point of $BD$ & $AC$.

$\therefore BO$ is median of $ABC$ also $OD$ is median

of $ADC$

$\therefore Ar\left(△AOD\right)=Ar\left(△DOC\right)$...(1)

Similarly $OP$ is median of $AOC$

$\therefore Ar\left(△AOP\right)=A\left(△POC\right)$...(2)

Substracting (2) From (1)

$Ar\left(△AOD\right)-Ar\left(△AOP\right)-Ar\left(△DOC\right)-Ar\left(△POC\right)$

$Ar\left(△APD\right)=Ar\left(△CPD\right)$

$\therefore$ Hence proved