$ABCD$ is a parallelogram whose diagonals intersect at $O$. If $P$ is any point on $BO$, prove that $ar(\triangle ABP) = ar(\triangle CBP)$.

Given:

$ABCD$ is a parallelogram whose diagonals intersect at $O$.

$P$ is a point on $BO$.

To do:

We have to prove that $ar(\triangle ABP) = ar(\triangle CBP)$.

Solution:

Join $AP$ and $CP$.

$O$ is the mid point of $AC$

This implies,

$PO$ is the median of $\triangle APC$.

Therefore,

$ar(\triangle APO) = ar(\triangle CPO)$......…(i)

Similarly,

$BO$ is the median of $\triangle ABC$

This implies,

$ar(\triangle ABO) = ar(\triangle BCO)$......…(ii)

Subtracting (i) from (ii), we get,

$ar(\triangle ABO) - ar(\triangle APO) = ar(\triangle BCO) - ar(\triangle CPO)$

$ar(\triangle ABP) = ar(\triangle CBP)$

Hence proved. 

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Updated on: 10-Oct-2022

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