Question

In $△ABC,D$ is the mid-point of $AB$ and $P$ is any point on $BC$. $CQ||PD$ meets $AB$ is $Q$, then prove that area $\left(△BPQ\right)=\frac{1}{2}$ area $\left(△ABC\right)$

Answer 1

Given that:- In $△$ABC in Which D is the mid point of $AB,PD||CQ$

Construction:- Join $C$ to $D$ and $Q$ to $P$.

Proof:- $CD$ is median of the $△$ $ABC$.

$\therefore ar(△BCD)=ar(△DAC)$

In the figure.

$△PDC$ and $△PDQ$ are on the same base $PQ$ and between the same paralels $PD$ and $CQ$.

$\therefore ar(△PDC)=ar(△PDQ)$

$\Rightarrow ar\left(BCD\right)=\frac{1}{2}ar\left(ABC\right)$

$\Rightarrow ar\left(BPD\right)+ar\left(PDC\right)=\frac{1}{2}ar\left(ABC\right)$

$\Rightarrow ar\left(BPD\right)+ar\left(PDQ\right)=\frac{1}{2}ar\left(ABC\right)$

Hence,

$ar\left(BPQ\right)=\frac{1}{2}ar\left(ABC\right)$

Hence Proved.