Question

ΔABC and ΔDBC are on the same base BC and on opposite sides of BC. If O is the point of intersection of BC and AD, prove that $\frac{ar(∆ABC)}{ar(∆DBC)}=\frac{AO}{DO}.$

Answer 1

Given: ΔABC and ΔDBC are on the same base BC and AD intersects BC at O.
To prove: $\frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{DBC}\right)}=\frac{\mathrm{AO}}{\mathrm{DO}}$
Construction : Draw AL⊥BC and DM⊥BC.

Proof: $\angle$ALO = $\angle$DMO = 90°
$\angle$AOL = $\angle$DOM (vertically-opposite angles)
∴ Δ ALO ∼Δ DMO [ By AA - similarity]
$\Rightarrow \frac{\mathrm{AL}}{\mathrm{DM}}=\frac{\mathrm{AO}}{\mathrm{DO}}$ ...................(i)
Therefore,
$\frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{DBC}\right)}=\frac{{\displaystyle \frac{1}{2}}\times \mathrm{BC}\times \mathrm{AL}}{{\displaystyle \frac{1}{2}}\times \mathrm{BC}\times \mathrm{DM}}=\frac{\mathrm{AL}}{\mathrm{DM}}=\frac{\mathrm{AO}}{\mathrm{DO}}[\mathrm{using}(\mathrm{i}\left)\right]$
Hence,$\frac{\mathrm{ar}\left(△\mathrm{ABC}\right)}{\mathrm{ar}\left(△\mathrm{DBC}\right)}=\frac{\mathrm{AO}}{\mathrm{DO}}$