Question

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

Hint: - The problem can be solved easily using the Intercept theorem.

Given $\vartriangle ABC$ in which $BD$ is the bisector of$\angle B$ and a line $PQ\parallel AC$meets $AB,BC$ and $BD$ at $P,Q$ and $R$ respectively.
For proof $\left( i \right)$
Considering small $\vartriangle BQP,BR$ is the bisector of$\angle B$.
Using the properties of similar triangles we have
$\therefore \dfrac{{BQ}}{{BP}} = \dfrac{{QR}}{{PR}} \\ \Rightarrow BQ.PR = BP.QR \\ \Rightarrow PR.BQ = QR.BP \\$ (Rearranging the terms amongst themselves)
For proof $\left( {ii} \right)$
In$\vartriangle ABC$ we have
$PQ\parallel AC$ (Given in the question)
$\Rightarrow \dfrac{{AB}}{{AP}} = \dfrac{{CB}}{{CQ}} \\ \Rightarrow AB \times CQ = CB \times AP \\ \Rightarrow AB \times CQ = BC \times AP \\$ (By using intercept theorem and then rearranging the terms)

Note: - The intercept theorem, also known as Thaleâ€™s theorem or basic proportionality theorem, is an important theorem in elementary geometry about the ratios of various line segments that are created if two intersecting lines are intercepted by a pair of parallels. In order to use the Intercept theorem, the first and foremost thing is to recognize the triangle correctly.