Question

In the given figure, if $PA=x,RC=y$ and $QB=z,$ then which one of the following is correct$?$

• A. $2y=x+z$
• B. $4y=x+z$
• C. $xy+yz=xz$
• D. $xy+xz=yz$

Answer 1

The correct option is C $xy+yz=xz$

Given, $PA=x,RC=y,QB=z$
$△BAP\sim △BCR$ ($\therefore$ $\angle$S is common and $\angle BAP=\angle BCR={90}^{\circ }$)
$\therefore$ $\frac{RC}{PA}=\frac{BC}{AB}$

$\Rightarrow \frac{BC}{AB}=\frac{y}{x}$ ....(i)
Also, $△ABQ\sim △ACR$ ($\angle A$ common, $\angle ABQ=\angle ACR={90}^{\circ }$)
$\therefore$ $\frac{RC}{BQ}=\frac{AC}{AB}$

$\Rightarrow \frac{AB-BC}{AB}=\frac{y}{z}$

$\Rightarrow 1-\frac{BC}{AB}=\frac{y}{z}$ ....(ii)
From (i) and (ii), we have
$1-\frac{y}{x}=\frac{y}{z}$

$\Rightarrow \frac{x-y}{x}=\frac{y}{z}$

$\Rightarrow xy=xz-yz$

$\Rightarrow xy+yz=xz$