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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}} $
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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.
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