Question

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

Answer 1

Given: ΔABC and ΔDBC are on the same base BC. AD and BC intersect at O.

Prove that:

Construction: Draw and .

Now, in ΔALO and ΔDMO, we have

$\therefore \frac{\mathrm{AL}}{\mathrm{DM}}=\frac{\mathrm{AO}}{\mathrm{DO}}$ (Corresponding sides are proportional)

$$\begin{gathered} \frac{\mathrm{Ar}\left(∆\mathrm{ABC}\right)}{\mathrm{Ar}\left(∆\mathrm{BCD}\right)}=\frac{{\displaystyle \frac{1}{2}}\times \mathrm{BC}\times \mathrm{AL}}{{\displaystyle \frac{1}{2}}\times \mathrm{BC}\times \mathrm{DM}} \\ =\frac{{\displaystyle \mathrm{AL}}}{\mathrm{DM}} \\ =\frac{\mathrm{AO}}{\mathrm{DO}} \end{gathered}$$