In figure, ABC and DBC are two triangles on the same base B C. If…

Question

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

✓

Answer

To prove : $\frac{\operatorname{ar}(\triangle ABC )}{\operatorname{ar}(\triangle D B C)}=\frac{ AO }{ DO }$
We know that,
Area $=\frac{1}{2} \times$ base $\times$ height
Area of $\triangle ABC =\frac{1}{2} \times BC \times AE\quad \quad \ldots \ldots(i)$
In $\triangle DBC$,
Area of $\triangle DBC =\frac{1}{2} \times BC \times DF\quad \quad \ldots \ldots(ii)$
Dividing (ii) by (i),
$\Rightarrow \frac{\text { Area of } \triangle ABC }{\text { Area of } \triangle DBC }=\frac{\frac{1}{2} \times BC \times AE }{\frac{1}{2} \times BC \times DF }$
$\Rightarrow \frac{\text { Area of } \triangle ABC }{\text { Area of } \triangle DBC }=\frac{ AE }{ DF }\quad \quad \ldots \ldots(iii)$
In $\triangle AOE$ and $\triangle DOF$
$\Rightarrow \angle AEO =\angle DFO$ (both right angles)
$\angle AOE =\angle DOF$ (Vertically opposite)
$\therefore \quad \triangle AOE \sim \triangle DOF$ by AAsimilarity
Hence, $\frac{ AE }{ DF }=\frac{ AO }{ DO }$($\because$ corresponding sides will be in same ratio)
Putting in (iii), we get
$\Rightarrow \frac{\text { Area } \triangle ABC }{\text { Area } \triangle DBC }=\frac{ AE }{ DF }=\frac{ AO }{ DO }$
$\therefore \frac{\text { Area } \triangle ABC }{\text { Area } \triangle DBC }=\frac{ AO }{ DO }$, Hence proved