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Triangles — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsTriangles3 Marks
Question
In the given figure, $\triangle\text{ABC}$ and $\triangle\text{DBC}$ are on the same base BC. If AD and BC intersect at O, prove that $\therefore\frac{\text{Area}(\triangle\text{ABC)}}{\text{Area}(\triangle\text{DBC})}=\frac{\text{AO}}{\text{DO}}$
✓

Answer

Given: $\triangle\text{ABC}$ and $\triangle\text{DBC}$ are on the same base BC. AD and BC intersect at O.
Prove that: $\therefore\frac{\text{Area}(\triangle\text{ABC)}}{\text{Area}(\triangle\text{DBC})}=\frac{\text{AO}}{\text{DO}}$
Construction: Draw $\text{AL}\perp\text{BC and }\text{DM}\perp\text{BC}.$
Now, in $\triangle\text{ALO}$ and $\triangle\text{DMO,}$ we heve
$\angle\text{ALO}=\angle\text{DMO}=90^\circ$
$\angle\text{AOL}=\angle\text{DOM}$ (vertically opposite angles)
Therefore $\triangle\text{ALO}\sim\triangle\text{DMO}$
$\therefore\frac{\text{AL}}{\text{DM}}=\frac{\text{AO}}{\text{DO}}$ (Corresponding sides are proportional)
$\therefore\frac{\text{Ar}(\triangle\text{ABC)}}{\text{Ar}(\triangle\text{BCD})}=\frac{\frac{1}{2}\text{BC}\times\text{AL}}{\frac{1}{2}\text{BC}\times\text{DM}}$
$=\frac{\text{AL}}{\text{DM}}$
$=\frac{\text{AO}}{\text{DO}}$

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