Question

In $\Delta ABC,D$ is the mid-point of side AB and P is any point on side BC. If line segment CQ$\parallel PD$ meets side AB at Q , then prove that ar$\left(\Delta BPQ\right)=\frac{1}{2}ar\left(\Delta ABC\right)$

Answer 1

Given: $D$ is the mid-point of $AB$ and $P$ is any point on $BC$ of $\Delta ABC,CQ||PD$ meets $AB$ in $Q$, To

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

Const: Join $CD$.

Proof: Since median of a triangle divides it into two triangles of equal area, so we have

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

Since, triangles on the same base and between the same parallels are equal in area, se we have

ar$\left(\Delta DPQ\right)=ar\left(\Delta DPC\right)$...........(2)

[$\because$ Triangle $DPQ$ and $DPC$ are on the same base $DP$ and between the same parallels $DP$ and $CQ$ ]

$ar\left(\Delta DPQ\right)+ar\left(\Delta DPB\right)$ $=ar\left(\Delta DPC\right)+ar\left(\Delta DPB\right)$

Hence, ar$\left(\Delta BPQ\right)=ar\left(\Delta BCD\right)=\frac{1}{2}ar\left(\Delta ABC\right)$ [Using $1$]