Question

Two triangles ABC and DBC lie on the same side of the base BC. From a point P on BC, PQ||AB and PR||BD are drawn. They meet AC in Q and DC in R respectively. Prove that QR||AD.

Answer 1

Given Two triangles ABC and DBC lie on the same side of the base BC. Points P,Q and R are points on BC,AC and CD respectively such that PR||BD and PQ||AB.

To prove QR||AD

Proof In $△ABC$, we have

$PQ||AB$

$\therefore$ $\frac{CP}{PB}=\frac{CQ}{QA}$........(i) [By Basic proportionality Theorem]

In $△BCD$, we have

$PR||BD$

$\therefore$ $\frac{CP}{PB}=\frac{CR}{RD}$........(ii) [By Thale's Theorem]

From (i) and (ii), we have

$\frac{CQ}{QA}=\frac{CR}{RD}$

Thus, in $△ACD$, Q and R are points on AC and CD respectively such that

$\frac{CQ}{QA}=\frac{CR}{RD}$

$\Rightarrow$ $QR||AD$ [By the converse of Basic Proportionality Theorem]