(a) In the given figure, PA, QB and RC are perpendicular to AC. If PA = x units, QB = y units and RC = z units, prove that 1/x+1/z=1/y.

\(\displaystyle \Delta \mathrm{ABQ} \sim \Delta \mathrm{ACR}\)

\(\displaystyle \frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{QB}}{\mathrm{RC}}=\frac{\mathrm{y}}{\mathrm{z}}\)

Similarly, \(\displaystyle \Delta \mathrm{CBQ} \sim \Delta \mathrm{CAP}\)

\(\displaystyle \frac{\mathrm{BC}}{\mathrm{AC}}=\frac{\mathrm{QB}}{\mathrm{PA}}=\frac{\mathrm{y}}{\mathrm{x}}\)

On adding (i) & (ii), we get

\[\begin{aligned} & \frac{A B}{A C}+\frac{B C}{A C}=\frac{y}{z}+\frac{y}{x} \\ & \frac{A B+B C}{A C}=y\left(\frac{1}{z}+\frac{1}{x}\right) \\ & \frac{A C}{A C}=y\left(\frac{1}{z}+\frac{1}{x}\right) \\ & \therefore \frac{1}{x}+\frac{1}{z}=\frac{1}{y} \end{aligned} \]

In \(\displaystyle \triangle \mathrm{ABD}\) and \(\displaystyle \triangle \mathrm{PQM}\)

\[\begin{aligned} & \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AD}}{\mathrm{PM}} \text { (given) } \\ & \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{2 \mathrm{BD}}{2 \mathrm{QM}}=\frac{\mathrm{AD}}{\mathrm{PM}} \\ & \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}=\frac{\mathrm{AD}}{\mathrm{PM}} \\ & \therefore \triangle \mathrm{ABD} \sim \triangle \mathrm{PQM} \\ & \therefore \angle \mathrm{~B}=\angle \mathrm{Q} \\ & \mathrm{In} \triangle \mathrm{ABC} \text { and } \triangle \mathrm{PQR} \\ & \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{QR}} \text { and } \angle \mathrm{B}=\angle \mathrm{Q} \\ & \triangle \mathrm{ABC} \sim \triangle \mathrm{PQR} \end{aligned} \]