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Triangles — Maths STD 10 — Question

Gujarat BoardEnglish MediumSTD 10MathsTriangles3 Marks
Question
In a $\triangle\text{ABC, D}$ and $E$ are points on the sides $AB$ and $AC$ respectively. For the following cases show that $DE || BC:$
$AD = 5.7\ cm, BD = 9.5\ cm, AE = 3.3\ cm$ and $EC = 5.5\ cm.$
✓

Answer


We have,
$DE || BC$
We have, $AD = 5.7\ cm, BD = 9.5\ cm, AE = 3.3\ cm$ and $EC = 5.5\ cm$
Now $\frac{\text{AD}}{\text{BD}}=\frac{5.7}{9.5}=\frac{57}{95}$
$\Rightarrow\frac{\text{AD}}{\text{BD}}=\frac{3}{5}$
And $\frac{\text{AE}}{\text{EC}}=\frac{3.3}{5.5}=\frac{33}{55}$
$\Rightarrow\frac{\text{AE}}{\text{EC}}=\frac{3}{5}$
Thus $DE$ divides sides $AB$ and $AC$ of $\triangle\text{ABC}$ in the same ratio.Therefore, by the converse of basic proportionality theorem. We have $DE || BC.$

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