Question

In the Figure, PA, QB and RC are perpendicular to AC. Prove that $\frac{1}{x}+\frac{1}{z}=\frac{1}{y}$.

Answer 1

Consider $△CAP$ and $△CBQ$

$\angle CAP=\angle CBQ={90}^{\circ }$

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

So by $AA$, $△CAP\sim △CBQ$

Hence $\frac{BQ}{AP}=\frac{BC}{AC}$

$\frac{y}{x}=\frac{BC}{AC}$ (1)

Now consider $△ACR$ and $△ABQ$

$\angle ACR=\angle ABQ={90}^{\circ }$

$\angle QAB=\angle RAC$ (common angle)

So by $AA$, $△ACR\sim △ABQ$

Hence $\frac{BQ}{CR}=\frac{AB}{AC}$

$\frac{y}{z}=\frac{AB}{AC}$ (2)

Adding (1) and (2)

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

$y(\frac{1}{x}+\frac{1}{z})=\frac{BC+AB}{AC}$

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

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

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

Hence proved.