Question
In the given figure, $\triangle\text{ABC}$ and $\triangle\text{DBC}$ have the same base BC. If AD and BC intersect at O, prove that $\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DBC})}=\frac{\text{AO}}{\text{DO}}.$
✓
Answer
Construction: Draw $\text{AX}\perp\text{CO}$ and $\text{DY}\perp\text{BO}.$
As,
$\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DBC})}=\frac{\frac{1}{2}\times\text{AX}\times\text{BC}}{\frac{1}{2}\times\text{DY}\times\text{BC}}$
$\Rightarrow\frac{\text{ar}(\triangle\text{ABC})}{\text{ar}(\triangle\text{DBC})}=\frac{\text{AX}}{\text{DY}}\dots(\text{i})$
In $\triangle\text{ABC}$ and $\triangle\text{DBC},\angle\text{AXY}=\angle\text{DYO}=90^\circ(\text{By construction})$
$\angle\text{AOX}=\angle\text{DOY}(\text{Vertically opposite angles})$$\therefore\triangle\text{AXY}\sim\triangle\text{DYO }(\text{By AA criterion})$
This completes the proof.
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