Question

In the given figure, ∆ABC and ∆DBC have the same base BC. If AD and BC intersect at O, prove that $\frac{\mathrm{ar}(∆ABC)}{\mathrm{ar}(∆DBC)}=\frac{AO}{DO}.$

Answer 1

$$\begin{gathered} \mathrm{Construction}:\mathrm{Draw}\mathrm{AX}\perp \mathrm{CO}\mathrm{and}\mathrm{DY}\perp \mathrm{BO}. \\ \mathrm{As}, \\ \frac{\mathrm{ar}(△\mathrm{ABC})}{\mathrm{ar}(△\mathrm{DBC})}=\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times \mathrm{AX}\times \mathrm{BC}}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\times \mathrm{DY}\times \mathrm{BC}} \\ \Rightarrow \frac{\mathrm{ar}(△\mathrm{ABC})}{\mathrm{ar}(△\mathrm{DBC})}=\frac{\mathrm{AX}}{\mathrm{DY}}...\left(\mathrm{i}\right) \\ \mathrm{In}△\mathrm{ABC}\mathrm{and}△\mathrm{DBC},\angle \mathrm{AXY}=\angle \mathrm{DYO}=90°(\mathrm{By}\mathrm{construction})\angle \mathrm{AOX}=\angle \mathrm{DOY}(\mathrm{Vertically}\mathrm{opposite}\mathrm{angles})\therefore △\mathrm{AXO}\sim △\mathrm{DYO}(\mathrm{By}\mathrm{AA}\mathrm{criterion})\therefore \frac{\mathrm{AX}}{\mathrm{DY}}=\frac{\mathrm{AO}}{\mathrm{DO}}(\mathrm{Thales}\text{'}\mathrm{s}\mathrm{theorem})...(\mathrm{ii})\mathrm{From}(\mathrm{i})\mathrm{and}(\mathrm{ii}),\mathrm{we}\mathrm{have}:\frac{\mathrm{ar}(△\mathrm{ABC})}{\mathrm{ar}(△\mathrm{DBC})}=\frac{\mathrm{AX}}{\mathrm{DY}}=\frac{\mathrm{AO}}{\mathrm{DO}}\mathrm{or},\frac{\mathrm{ar}(△\mathrm{ABC})}{\mathrm{ar}(△\mathrm{DBC})}=\frac{\mathrm{AO}}{\mathrm{DO}} \end{gathered}$$
This completes the proof.