Question

In triangle ABC,D is the midpoint of AB,P is any point on BC, CG||PD meets BA produced at Q. Show that $ar\left(\Delta BPQ\right)=\frac{1}{2}$ ar($\Delta$ ABC).

Answer 1

In$△ABC,D$ is the mid-point of $AB$ and $D$ is any point on $BC.$

If$CQ||PD$ meets $AB$ in $Q.$

$ThenIn△ABC,$we have to prove that $BPQ=\frac{1}{2}ar\left(ABC\right)$

Construct $DC.$

Since $D$ is the mid point of $AB$ in $△BC,CD$ is the median.

$ar\left(△BCD\right)=\frac{1}{2}ar\left(△ABC\right)-\left(i\right)$

Since $△PDQ\&△PDC$ are in the same $PD$ and between the same parallel lines $PD\&QC.$

$\therefore ar\left(△PDQ\right)=ar\left(△PDC\right)-\left(ii\right)$

From$\left(i\right)\&\left(ii\right)$

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

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

$\{areaof△PDC=PDQ\}$

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