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In the given figure PA, QB and RC are each perpendicular to AC. If AP = $x$, BQ = $y$ and CR = $z$, then prove that $\frac{1}{x} + \frac{1}{z} = \frac{1}{y}$.
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Sol. $\triangle PAC \sim \triangle QBC$
$\therefore \frac{x}{y} = \frac{AC}{BC}$ or $\frac{y}{x} = \frac{BC}{AC}$ ----- (i)
$\triangle RCA \sim \triangle QBA$
$\therefore \frac{z}{y} = \frac{AC}{AB}$ or $\frac{y}{z} = \frac{AB}{AC}$ ----- (ii)
Adding (i) and (ii)
$\frac{y}{x} + \frac{y}{z} = \frac{BC+AB}{AC}$
$\Rightarrow y\left(\frac{1}{x} + \frac{1}{z}\right) = \frac{AC}{AC} = 1$
$\Rightarrow \frac{1}{x} + \frac{1}{z} = \frac{1}{y}$
$\therefore \frac{x}{y} = \frac{AC}{BC}$ or $\frac{y}{x} = \frac{BC}{AC}$ ----- (i)
$\triangle RCA \sim \triangle QBA$
$\therefore \frac{z}{y} = \frac{AC}{AB}$ or $\frac{y}{z} = \frac{AB}{AC}$ ----- (ii)
Adding (i) and (ii)
$\frac{y}{x} + \frac{y}{z} = \frac{BC+AB}{AC}$
$\Rightarrow y\left(\frac{1}{x} + \frac{1}{z}\right) = \frac{AC}{AC} = 1$
$\Rightarrow \frac{1}{x} + \frac{1}{z} = \frac{1}{y}$