In the given figure PA, QB and RC are each perpendicular to AC.

Q) In the given figure PA, QB and RC are each perpendicular to AC. If AP = x, BQ = y and CR = z, then prove that 1/x + 1/z = 1/y

Ans:

Step 1: In Δ APC and Δ BQC,

∠ PCA = ∠ QCB (common angle)

∠ PAC = ∠ QBC (90⁰)

By AA identity, Δ APC ~ Δ BQC

Hence, $\frac{QB}{PA} = \frac{BC}{AC}$

$\Rightarrow \frac{y}{x} = \frac{BC}{AC}$ ………….. (i)

Step 2: In Δ RCA and Δ QBA,

∠ RAC = ∠ QAB (common angle)

∠ RCA = ∠ QBA (90⁰)

By AA identity, Δ RCA ~ Δ QBA

Hence, $\frac{QB}{RC} = \frac{AB}{AC}$

$\Rightarrow \frac{y}{z} = \frac{AB}{AC}$ ………….. (ii)

By adding equations (i) and (ii), we get:

$\frac{y}{x} + \frac{y}{z} = \frac{BC}{AC} + \frac{AB}{AC}$

$\Rightarrow \frac{y}{x} + \frac{y}{z} = \frac{BC + AB}{AC}$

$\Rightarrow \frac{y z + y x}{x z} = \frac{AC}{AC} = 1$

$\Rightarrow y z + y x = x z$

Dividing by x y z on both sides, we get:

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

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

Hence Proved !

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