(a) PA, QB and RC are each perpendicular to AC. If AP =x, QB = z, RC = y, AB = a and BC = b, then prove that 1/x+1/y=1/z.

\(\displaystyle \Delta \mathrm{CQB} \sim \Delta \mathrm{CPA}\)

\[\begin{aligned} & \Rightarrow \frac{\mathrm{b}}{\mathrm{a}+\mathrm{b}}=\frac{\mathrm{z}}{\mathrm{x}}-\text { (i) } \\ & \text { Also } \Delta \mathrm{AQB} \sim \Delta \text { ARC } \\ & \Rightarrow \frac{\mathrm{a}}{\mathrm{a}+\mathrm{b}}=\frac{z}{y}-\text { (ii) } \\ & \text { from (i) and (ii) } \frac{\mathrm{z}}{\mathrm{x}}+\frac{\mathrm{z}}{\mathrm{y}}=\frac{\mathrm{a}+\mathrm{b}}{\mathrm{a}+\mathrm{b}}=1 \\ & \Rightarrow \frac{1}{\mathrm{x}}+\frac{1}{\mathrm{y}}=\frac{1}{\mathrm{z}} \end{aligned} \]

\[\begin{aligned} & \text { (i) } \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR} \\ & \Rightarrow \angle \mathrm{~A}=\angle \mathrm{P} \\ & \text { and } \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}} \\ & \Rightarrow \frac{2 A D}{2 P S}=\frac{A C}{P R} \end{aligned} \]

\[\Rightarrow \frac{A D}{P S}=\frac{A C}{P R} \text { and } \angle \mathrm{A}=\angle \mathrm{P} \]

Therefore \(\displaystyle \Delta \mathrm{ADC} \sim \Delta \mathrm{PSR}\)

\[\begin{aligned} & \text { (ii)Hence } \frac{A D}{P S}=\frac{A C}{P R} \\ & \Rightarrow \mathrm{AD} \times \mathrm{PR}=\mathrm{AC} \times \mathrm{PS} \end{aligned} \]