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PA, QB and RC are each perpendicular to AC. If AP = $x$, QB = $z$, RC = $Y$, AB = $a$ and BC = $b$, then prove that $\frac{1}{x} + \frac{1}{y} = \frac{1}{z}$
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(a)$\triangle CQB \sim \triangle CPA$
$\Rightarrow \frac{b}{a + b} = \frac{z}{x}$ (i)
Also $\triangle AQB \sim \triangle ARC$
$\Rightarrow \frac{a}{a + b} = \frac{z}{y}$ (ii)
from (i) and (ii) $\frac{z}{x} + \frac{z}{y} = \frac{a + b}{a + b} = 1$
$\Rightarrow \frac{1}{x} + \frac{1}{y} = \frac{1}{z}$
$\Rightarrow \frac{b}{a + b} = \frac{z}{x}$ (i)
Also $\triangle AQB \sim \triangle ARC$
$\Rightarrow \frac{a}{a + b} = \frac{z}{y}$ (ii)
from (i) and (ii) $\frac{z}{x} + \frac{z}{y} = \frac{a + b}{a + b} = 1$
$\Rightarrow \frac{1}{x} + \frac{1}{y} = \frac{1}{z}$