Question

State the whether the given statement is true or false

In the given figure $QR$ is parallel to $AB$ and $DR$ is parallel to $QB$ then $P{Q}^2=PD\times PA$.

• A. True
• B. False

Answer 1

The correct option is A True

$QR||AB$ & $DR||QB$

In $\Delta PQR$ & $\Delta PAB$
i)$\angle P$ is common
ii)$\angle PQR=\angle PAB(\because \angle PQR=\angle PAB=alt.int.\angle sofQR||ABandPAistransversal)$

$\because \Delta PQR$ $\sim \Delta PAB$ (by $AA$ similarity).

$\therefore \frac{PQ}{PA}=\frac{PR}{PB}=\frac{QR}{AB}$ _____ (i)

Similarly, In $\Delta PDR$ and $\Delta PQB$

i)$\angle P$ is common
ii)$\angle PDR=\angle PQB=alt.int.\angle sofDR||QBandPQistransversal$

$\Delta PDR$ $\sim \Delta PQB$ (by $AA$ similarity)

$\therefore \frac{PD}{PQ}=\frac{PR}{PB}=\frac{DR}{QB}$ ________ (ii)

Comparing equation (i) and equation (ii), we get

$\Rightarrow \frac{PQ}{PA}=\frac{PD}{PQ}$

$\Rightarrow P{Q}^2=PA\times PD$ (proved)