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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 $DE || BC:$
$AB = 12\ cm, AD = 8\ cm, AE = 12\ cm$ and $AC = 18\ cm.$
✓

Answer


$\text{AB} = 12\text{cm}, \text{AD} = 12\text{cm}\ \text{and AC} = 18\text{cm.}$
$\therefore\text{AD}=\text{AB}-\text{AD}$
$=12-8$
$\Rightarrow\text{DB}=4\text{cm}$
$\text{And, EC}=\text{AC}-\text{AE}$
$=18-12$
$\Rightarrow\text{EC}=6\text{cm}$
$\text{Now},\frac{\text{AD}}{\text{DB}}=\frac{8}{4}=\frac{2}{1}\ \ [\because\text{DB}=4\text{cm}]$
$\text{And},\frac{\text{AE}}{\text{EC}}=\frac{12}{6}=\frac{2}{1}\ \ [\because\text{EC}=6\text{cm}]$
$\Rightarrow\frac{\text{AD}}{\text{DB}}=\frac{\text{AE}}{\text{EC}}$
Thus, $DE$ divides sides $AB$ and $AC$ of $\triangle\text{ABC}$ in the same ratio.
Therefore, by the converse of basic proportionality theorem.

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