Question

In fig 1.40, $△\mathrm{ABC}$ is right-angled at B.
D is any point on AB and seg DE$\perp$ seg AC.
If AD=6 cm, AB=12 cm and AC=18 cm, find AE.

Answer 1

$$\begin{gathered} \text{In}△\mathrm{ABC}\text{and}△\mathrm{AED}, \\ \angle \mathrm{ABC}\cong \angle \mathrm{AED}(90°\mathrm{each}) \\ \angle \mathrm{BAC}\cong \angle \mathrm{EAD}(\mathrm{Common}\mathrm{angle}) \\ \mathrm{Therefore},\mathrm{by}\text{the}\mathrm{AA}\mathrm{similarity}\mathrm{test}, \\ △\mathrm{ABC}~△\mathrm{AED}. \\ \mathrm{Hence},\mathrm{by}\mathrm{the}\mathrm{property}\mathrm{of}\mathrm{similarity}, \\ \frac{\mathrm{AB}}{\mathrm{AE}}=\frac{\mathrm{AC}}{\mathrm{AD}} \\ \Rightarrow \frac{12}{\mathrm{AE}}=\frac{18}{6} \\ \Rightarrow \mathrm{AE}=\frac{12\times 6}{18}=4\mathrm{cm} \end{gathered}$$