Question

In a $△$ $ABC,D$ and $E$ are points on the sides AB and AC respectively. For each of the following cases show that $DE$ $\parallel$ $BC$:

(i) $AB=12cm,AD=8cm,AE=12cmandAC=18cm$.

(ii) $AB=5.6cm,AD=1.4cm,AC=7.2cmandAE=1.8cm$.

(iii) $AB=10.8cm,BD=4.5cm,AC=4.8cmandAE=2.8cm$.

(iv) $AD=5.7cm,BD=9.5cm,AE=3.3cmandEC=5.5cm.$

Answer 1

Here, using the corollary of basic proportionality theorem which states that if a line passing through the two sides of the triangle cuts it proportionally, then the line is parallel to the third side. So,

$\left(i\right)$ $\frac{AB}{AD}=\frac{12}{8}=\frac{3}{2}$

$\frac{AC}{AE}=\frac{18}{12}=\frac{3}{2}$

$\therefore$ $\frac{AB}{AD}=\frac{AC}{AE}$

Thus, as $DE$ cuts the sides $AB$ and $AC$ proportionally, so

$DE\parallel BC$.

$\therefore$ $DE\parallel BC$

$\left(ii\right)$ $\frac{AB}{AD}=\frac{5.6}{1.4}=4$

$\frac{AC}{AE}=\frac{7.2}{1.8}=4$

$\therefore$ $\frac{AB}{AD}=\frac{AC}{AE}$

Thus, as $DE$ cuts the sides $AB$ and $AC$ proportionally, so

$DE\parallel BC$.

$\therefore$ $DE\parallel BC$

$\left(iii\right)$ $\frac{AD}{BD}=\frac{10.8-4.5}{4.5}=\frac{6.3}{4.5}=\frac{7}{5}$

$\frac{AE}{EC}=\frac{2.8}{4.8-2.8}=\frac{2.8}{2}=\frac{7}{5}$

$\therefore$ $\frac{AD}{BD}=\frac{AE}{EC}$

Thus, as $DE$ cuts the sides $AB$ and $AC$ proportionally, so

$DE\parallel BC$.

$\therefore$ $DE\parallel BC$

$\left(iv\right)$ $\frac{AD}{BD}=\frac{5.7}{9.5}=\frac{3}{5}$

$\frac{AE}{EC}=\frac{3.3}{5.5}=\frac{3}{5}$

$\therefore$ $\frac{AD}{BD}=\frac{AE}{EC}$

Thus, as $DE$ cuts the sides $AB$ and $AC$ proportionally, so

$DE\parallel BC$.

$\therefore$ $DE\parallel BC$