Question

In ∆ABC, AL and CM are the perpendiculars from the vertices A and C to BC and AB respectively. If AL and CM intersect at O, prove that:

(i) ∆OMA ∼ ∆OLC
(ii) $\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OM}}{\mathrm{OL}}$

Answer 1

$$\begin{gathered} \left(\mathrm{i}\right).\mathrm{In}∆\mathrm{OMA}\mathrm{and}∆\mathrm{OLC}, \\ \angle \mathrm{AOM}=\angle \mathrm{COL}\left[\mathrm{Vertically}\mathrm{opposite}\mathrm{angles}\right] \\ \angle \mathrm{OMA}=\angle \mathrm{OLC}\left[90°\mathrm{each}\right] \\ \Rightarrow ∆\mathrm{OMA}~∆\mathrm{OLC}\left[\mathrm{AA}\mathrm{similarity}\right] \\ \left(\mathrm{ii}\right).\mathrm{Since}∆\mathrm{OMA}~∆\mathrm{OLC}\mathrm{by}\mathrm{AA}\mathrm{similarity},\mathrm{then} \\ \frac{\mathrm{OM}}{\mathrm{OL}}=\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{MA}}{\mathrm{LC}}\left[\mathrm{Corresponding}\mathrm{sides}\mathrm{of}\mathrm{similar}\mathrm{triangles}\mathrm{are}\mathrm{proportional}\right] \\ \Rightarrow \frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OM}}{\mathrm{OL}} \end{gathered}$$