Question

$D$ and $E$ are respectively the points on the sides $AB$ and $AC$ of a $△ABC$ such that $AB=12\text{cm},$ $AD=8\text{cm},$ $AE=12\text{cm}$ and $AC=18\text{cm}$, then

• A. $DE\parallel BD$
• B. $DE\parallel BC$
• C. $AD\parallel BD$
• D. $AD\parallel CD$

Answer 1

The correct option is B $DE\parallel BC$

If $\frac{AD}{BD}=\frac{AE}{CE}$ then by converse of basic proportionality theorem we can say that $DE||BC.$

$\begin{array}{cc}\frac{AD}{BD}& =\frac{AE}{CE}\dots \left(i\right)\\ \frac{BD}{AD}& =\frac{CE}{AE}\\ \frac{BD}{AD}+1& =\frac{CE}{AE}+1\\ \frac{BD+AD}{AD}& =\frac{CE+AE}{AE}\\ \frac{AB}{AD}& =\frac{AC}{AE}\\ \frac{AD}{AB}& =\frac{AE}{AC}\dots \left(ii\right)\end{array}$

So, the above calculation implies that instead of an equation $\left(i\right)$ if we prove that equation $\left(ii\right)$ is true then also we can say that $DE||BC.$

$\begin{array}{cc}\frac{AD}{AB}& =\frac{AE}{AC}\\ \frac{8}{12}& =\frac{12}{18}\\ \frac{2}{3}& =\frac{2}{3}\left[\text{Which is true}\right]\end{array}$

Hence, proved $DE||BC.$