In the given figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that$ \frac{\operatorname{ar}(\triangle \mathbf{A B C})}{\operatorname{ar}(\triangle \mathbf{D B C})}=\frac{\mathbf{A O}}{\mathbf{D O}} $
"

Given:

ABC and DBC are two triangles on the same base BC.

AD intersects BC at O

To do:

We have to show that \( \frac{\operatorname{ar}(\triangle \mathbf{A B C})}{\operatorname{ar}(\triangle \mathbf{D B C})}=\frac{\mathbf{A O}}{\mathbf{D O}} \)

Solution:

Draw $AM \perp BC$ and $DN \perp BC$

In $\triangle AOM$ and $\triangle DON$,

$\angle \mathrm{AOM}=\angle \mathrm{DON}$             (Vertically opposite angles)

$\angle \mathrm{AMO}=\angle \mathrm{DNO}=90^o$

Therefore, by AA similarity,

$\Delta \mathrm{AOM} \sim \Delta \mathrm{DON}$

This implies,

$\frac{\mathrm{AM}}{\mathrm{DN}}=\frac{\mathrm{AO}}{\mathrm{DO}}$          (CPCT)

Therefore,

$\frac{\operatorname{ar} \Delta \mathrm{ABC}}{\operatorname{ar} \Delta \mathrm{DBC}}=\frac{\frac{1}{2} \times \mathrm{BC} \times \mathrm{AM}}{\frac{1}{2} \times \mathrm{BC} \times \mathrm{DN}}$

$=\frac{\mathrm{AM}}{\mathrm{DN}}$

$\frac{\operatorname{ar} \triangle \mathrm{ABC}}{\operatorname{ar} \triangle \mathrm{DBC}}=\frac{\mathrm{AO}}{\mathrm{DO}}$           (Since $\frac{\mathrm{AM}}{\mathrm{DN}}=\frac{\mathrm{AO}}{\mathrm{DO}}$)

Hence proved.