Question

In fig., if PQ $\parallel$ BC and PR $\parallel$CD, then, $\frac{AB}{AD}$=

• A. $\frac{AC}{PC}$
• B. $\frac{QB}{QA}$
• C. $\frac{QB}{RD}$
• D. None of the above

Answer 1

The correct option is C $\frac{QB}{RD}$

In $\Delta ABC$, we have

$PQ\parallel CB$ [Given]

Therefore, by basic proportionality theorem, we have

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

Adding 1 on both sides we get,

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

$\Rightarrow \frac{AB}{QB}=\frac{AC}{PC}........\left(1\right)$

In $\Delta ACD$, we have

$PR\parallel CD$ [Given]

Therefore, by basic proportionality theorem, we have

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

Adding 1 on both sides we get,

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

$\Rightarrow \frac{AC}{PC}=\frac{AD}{RD}........\left(2\right)$

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

$\frac{AB}{QB}$ = $\frac{AD}{RD}$

$\Rightarrow \frac{AB}{AD}$ = $\frac{QB}{RD}$