Question

In the adjoining figure, AD is a median of ∆ABC. If BL and CM are drawn perpendiculars on AD and AD produced, prove that BL = CM.

Answer 1

$$\begin{gathered} \mathrm{In}△\mathrm{BLD}\mathrm{and}△\mathrm{CMD},\mathrm{we}\mathrm{have}: \\ \angle \mathrm{BLD}=\angle \mathrm{CMD}(90°\mathrm{each}) \\ \mathrm{BD}=\mathrm{CD}(\mathrm{AD}\mathrm{is}\mathrm{median},\mathrm{i}.\mathrm{e}.,\mathrm{D}\mathrm{is}\mathrm{the}\mathrm{midpoint}\mathrm{of}\mathrm{BC}) \\ \angle \mathrm{BDL}=\angle \mathrm{CDM}(\mathrm{Vertically}\mathrm{opposite}\mathrm{angle}) \\ △\mathrm{BLD}\cong △\mathrm{CMD}(\mathrm{AAS}\mathrm{criterion}) \\ \therefore \mathrm{BL}=\mathrm{CM}(\mathrm{Corresponding}\mathrm{sides}\mathrm{of}\mathrm{congruent}\mathrm{triangles}) \end{gathered}$$
Hence, proved.