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In figure below, $PA, QB$ and $RC$ are each perpendicular to $AC$. Prove that $\frac{1}{x}+\frac{1}{z}=\frac{1}{y}$.
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Academic Mathematics NCERT Class 10

Given:

$PA \perp AC, QB \perp AC$ and $RC \perp AC$.
To do:

We have to prove that $\frac{1}{x}+\frac{1}{z}=\frac{1}{y}$.

Solution:

$AP=x$ and $BQ=y$

Let $AB=a$ and $BC=b$

In $\triangle CQB$ and $\triangle CPA$,

$\angle QBC=\angle PAC=90^o$

$\angle QCB=\angle PCA$ (Common angle)

Therefore,

$\triangle CQB \sim\ \triangle CPA$ (By AA similarity)

This implies,

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

$\frac{y}{x}=\frac{b}{a+b}$.....(i)

In $\triangle AQB$ and $\triangle ARC$,

$\angle ABQ=\angle ACR=90^o$

$\angle BAQ=\angle CAR$ (Common angle)

Therefore,

$\triangle AQB \sim\ \triangle ARC$ (By AA similarity)

This implies,

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

$\frac{y}{z}=\frac{a}{a+b}$.....(ii)

Adding equations (i) and (ii), we get,

$\frac{y}{x}+\frac{y}{z}=\frac{b}{a+b}+\frac{a}{a+b}$

$\frac{yz+xy}{xz}=\frac{a+b}{a+b}$

$\frac{yz+xy}{xz}=1$

$yz+xy=xz$

Dividing both sides by $xyz$, we get,

$\frac{yz+xy}{xyz}=\frac{xz}{xyz}$

$\frac{yz}{xyz}+\frac{xy}{xyz}=\frac{xz}{xyz}$

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

Hence proved.

Updated on: 10-Oct-2022

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