Question

In the given figure, PA, OB and RC are each perpendicular to AC. Which of the relations hold true?

• A. $\frac{1}{x}+\frac{1}{y}<\frac{1}{z}$.
• B. $\frac{1}{x}+\frac{1}{z}=\frac{1}{y}$.
• C. $\frac{1}{x}+\frac{1}{z}>\frac{1}{y}$.
• D. $\frac{1}{x}-\frac{1}{y}=\frac{1}{z}$

Answer 1

Answer: (B)

$\frac{1}{x}+\frac{1}{z}=\frac{1}{y}$.

In $\Delta PAC$, we have $BQ\parallel AP$
$\Rightarrow \frac{BQ}{AP}=\frac{CB}{CA}[\therefore \Delta CBQ\sim \Delta CAP]byAAA$
$\Rightarrow \frac{y}{x}=\frac{CB}{CA}$.....(i)
In $\Delta ACR$, we have $BQ\parallel CR$
$\Rightarrow \frac{BQ}{CR}=\frac{AB}{AC}[\therefore \Delta ABQ\sim \Delta ACR]$ by AAA
$\Rightarrow \frac{y}{z}=\frac{AB}{AC}$......(ii)
Adding (i) and (ii), we get
$\frac{y}{x}+\frac{y}{z}=\frac{CB}{AC}+\frac{AB}{AC}$
$\Rightarrow \frac{y}{x}+\frac{y}{z}=\frac{AB+BC}{AC}$
$\Rightarrow \frac{y}{x}+\frac{y}{z}=\frac{AC}{AC}$
$\Rightarrow \frac{y}{x}+\frac{y}{z}=1$
$\Rightarrow \frac{1}{x}+\frac{1}{z}=\frac{1}{y}$.
hence Proved