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In the given figure, PA, QB and RC are perpendicular to AC. If $PA = x$ units, $QB = y$ units and $RC = z$ units, prove that $\frac{1}{x} + \frac{1}{z} = \frac{1}{y}$.
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Sol. $\triangle ABQ \sim \triangle ACR$
$\frac{AB}{AC} = \frac{QB}{RC} = \frac{y}{z}$ ... (i)
Similarly, $\triangle CBQ \sim \triangle CAP$
$\frac{BC}{AC} = \frac{QB}{PA} = \frac{y}{x}$ ... (ii)
On adding (i) & (ii), we get
$\frac{AB}{AC} + \frac{BC}{AC} = \frac{y}{z} + \frac{y}{x}$
$\frac{AB+BC}{AC} = y(\frac{1}{z} + \frac{1}{x})$
$\frac{AC}{AC} = y(\frac{1}{z} + \frac{1}{x})$
$1 = y(\frac{1}{z} + \frac{1}{x})$
$\therefore \frac{1}{y} = \frac{1}{x} + \frac{1}{z}$
$\frac{AB}{AC} = \frac{QB}{RC} = \frac{y}{z}$ ... (i)
Similarly, $\triangle CBQ \sim \triangle CAP$
$\frac{BC}{AC} = \frac{QB}{PA} = \frac{y}{x}$ ... (ii)
On adding (i) & (ii), we get
$\frac{AB}{AC} + \frac{BC}{AC} = \frac{y}{z} + \frac{y}{x}$
$\frac{AB+BC}{AC} = y(\frac{1}{z} + \frac{1}{x})$
$\frac{AC}{AC} = y(\frac{1}{z} + \frac{1}{x})$
$1 = y(\frac{1}{z} + \frac{1}{x})$
$\therefore \frac{1}{y} = \frac{1}{x} + \frac{1}{z}$