Question

In the given figure, PA, QB and RC are perpendicular to AC. If AP = x, QB = z, RC = y, AB = a and BC = b, show that $\frac{1}{x}+\frac{1}{y}=\frac{1}{z}.$

Answer 1

$$\begin{gathered} \mathrm{In}△PAC\mathrm{and}△QBC,\mathrm{we}\mathrm{have}: \\ \angle A=\angle B\left(\mathrm{Both}\mathrm{angles}\mathrm{are}90°\right) \\ \angle P=\angle Q\left(\mathrm{Correspond}\text{ing}\mathrm{angles}\right) \\ \mathrm{and} \\ \angle C=\angle C\left(\mathrm{Common}\mathrm{angles}\right) \\ \mathrm{Therefore},△PAC~△QBC \\ \frac{AP}{BQ}=\frac{AC}{BC} \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{x}{z}=\frac{a+b}{b} \\ \Rightarrow a+b=\frac{bx}{z}...\left(1\right) \end{gathered}$$

$$\begin{gathered} \mathrm{In}△RCA\mathrm{and}△QBA,\mathrm{we}\mathrm{have}: \\ \angle C=\angle B\left(\mathrm{Both}\mathrm{angles}\mathrm{are}90°\right) \\ \angle R=\angle Q\left(\mathrm{Correspond}\text{ing}\mathrm{angles}\right) \\ \mathrm{and} \\ \angle A=\angle A\left(\mathrm{Common}\mathrm{angles}\right) \\ \mathrm{Therefore},△RCA~△QBA \\ \frac{RC}{BQ}=\frac{AC}{AB} \\ \Rightarrow \frac{y}{z}=\frac{a+b}{a} \\ \Rightarrow a+b=\frac{ay}{z}...\left(2\right) \end{gathered}$$
From equation (1) and (2), we have:

$$\begin{gathered} \frac{bx}{z}=\frac{ay}{z} \\ \Rightarrow bx=ay \\ \Rightarrow \frac{a}{b}=\frac{x}{y}...\left(3\right) \\ \mathrm{Also}, \\ \frac{x}{z}=\frac{a+b}{b} \\ \Rightarrow \frac{x}{z}=\frac{a}{b}+1 \\ \mathrm{Using}\mathrm{the}\mathrm{value}\mathrm{of}\frac{a}{b}\mathrm{from}\mathrm{equation}\left(3\right),\mathrm{we}\mathrm{have}: \\ \Rightarrow \frac{x}{z}=\frac{x}{y}+1 \\ \mathrm{Dividing}\mathrm{both}\mathrm{sides}\mathrm{by}x,\mathrm{we}\mathrm{get}: \\ \frac{1}{z}=\frac{1}{y}+\frac{1}{x} \end{gathered}$$

$$\begin{gathered} \therefore \frac{1}{x}+\frac{1}{y}=\frac{1}{z} \\ \mathrm{This}\mathrm{completes}\mathrm{the}\mathrm{proof}. \end{gathered}$$