Question

Two poles of heights 'a' metres and 'b' metres are 'p' metres apart. Prove that the height 'h' drawn from the point of intersection N of the lines joining the top of each pole to the foot of the opposite pole is $\frac{ab}{a+b}$ metres.

Answer 1

$$\begin{gathered} \mathrm{In}△\mathrm{RBA}\text{and}△\mathrm{NTA}, \\ \angle \mathrm{RBA}=\angle \mathrm{NTA}[90°\mathrm{each}] \\ \angle \mathrm{RAB}=\angle \mathrm{NAT}[\mathrm{Common}\mathrm{angle}\text{s}] \\ \mathrm{Therefore},\mathrm{by}\mathrm{AA}\mathrm{similarity}\mathrm{test}, \\ △\mathrm{RBA}~△\mathrm{NTA} \\ \therefore \frac{\mathrm{RB}}{\mathrm{NT}}=\frac{\mathrm{AB}}{\mathrm{AT}} \\ \Rightarrow \frac{a}{h}=\frac{p}{x} \\ \Rightarrow a=\frac{hp}{x}...\left(1\right) \\ \mathrm{In}△\mathrm{SAB}\text{and}△\mathrm{NTB}, \\ \angle \mathrm{SAB}=\angle \mathrm{NTB}[90°\mathrm{each}] \\ \angle \mathrm{SBA}=\angle \mathrm{NBT}[\mathrm{Common}\mathrm{angle}\text{s}] \\ \mathrm{Therefore},\mathrm{by}\mathrm{AA}\mathrm{similarity}\mathrm{test}, \\ △\mathrm{SAB}~△\mathrm{NTB} \\ \therefore \frac{\mathrm{SA}}{\mathrm{NT}}=\frac{\mathrm{AB}}{\mathrm{TB}} \\ \Rightarrow \frac{b}{h}=\frac{p}{y} \\ \Rightarrow b=\frac{hp}{y}...\left(2\right) \\ \mathrm{From}\left(1\right)\&\left(2\right),\text{we get:} \\ ab=\frac{hp}{x}\times \frac{hp}{y} \\ ab=\frac{h^2{p}^2}{xy}...\left(3\right) \\ \mathrm{From}\left(1\right)\&\left(2\right),\text{we get:} \\ a+b=\frac{hp}{x}+\frac{hp}{y} \\ =\frac{hpy+hpx}{xy} \\ =hp\left(\frac{1}{x}+\frac{1}{y}\right)...\left(4\right) \\ \frac{ab}{a+b}=\frac{{\displaystyle \frac{h^2{p}^2}{xy}}}{hp\left(\frac{1}{x}+\frac{1}{y}\right)}[\mathrm{From}(3)\&(4\left)\right] \\ =\frac{hp}{x+y} \\ =\frac{hp}{p} \\ =h \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$