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

It is given that $P A, Q B$ and $R C$ are each perpendicular to $A C$.

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

In $\triangle P A C$ we have $B Q \| A P$

$\Rightarrow \frac{B Q}{A P}=\frac{C B}{C A}$

$\Rightarrow y x=C B C A$.....(1)

Now in $\triangle A C R$, we have $B Q \| C R$

$\Rightarrow \frac{B Q}{C R}=\frac{A B}{A C}$

$\Rightarrow \frac{y}{z}=\frac{A B}{A C}$....(2)

Adding (1) and (2) we have

$\frac{y}{x}+\frac{y}{z}=\frac{C B}{A C}+\frac{A B}{A C}$

$=\frac{A B+B C}{A C}$

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

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

Hence, $\frac{1}{x}+\frac{1}{z}=\frac{1}{y}$.