Question

In $△ABC$, the bisector of $\angle B$ meets AC at D. A line PQ||AC meets AB,BC and BD at P,Q and R respectively. Show that
(i) $PR.BQ=QR.BP$
(ii) $AB\times CQ=BC\times AP$

Answer 1

Given $△ABC$ in which BD is the bisector of $\angle B$ and a line PQ||AC meets AB,BC and BD at P,Q and R respectively.
Proof (i)

In $△BQP$, BR is the bisector of $\angle B$.

$\therefore$ $\frac{BQ}{BP}=\frac{QR}{PR}$

$\Rightarrow$ $BQ.PR=BP.QR$

$\Rightarrow$ $PR.BQ=QR.BP$ $\left[Henceproved\right]$

(ii) In $△ABC$, we have

$PQ||AC$ [Given]

$\Rightarrow$ $\frac{AB}{AP}=\frac{CB}{CQ}$ [By Thale's Theorem]

$\Rightarrow$ $AB\times CQ=BC.AP$ $\left[Henceproved\right]$