Question

In fig., If PQ $\parallel$ BC and PR $\parallel$CD. Prove that $\frac{AR}{AD}$= $\frac{AQ}{AB}$. [4 MARKS]

Answer 1

Concept : 1 Mark
Application : 1 Mark
Proof : 2 Marks

In $\Delta ABC$, we have

$PQ\parallel CB$ [Given]

Therefore, by basic proportionality theorem, we have

$\frac{AQ}{QB}$ = $\frac{AP}{PC}$

$\Rightarrow \frac{AQ}{AB}=\frac{AP}{AC}........\left(1\right)$

In $\Delta ACD$, we have

$PR\parallel CD$ [Given]

Therefore, by basic proportionality theorem, we have

$\frac{AP}{PC}$ = $\frac{AR}{RD}$

$\Rightarrow \frac{AP}{AC}$ = $\frac{AR}{AD}........\left(2\right)$

From (1) and (2), we obtain that

$\frac{AQ}{AB}$ = $\frac{AR}{AD}$

or $\frac{AR}{AD}$ = $\frac{AQ}{AB}$