Question

In the given figure, ABC is a triangle, DE is parallel to BC and $\frac{AD}{DB}=\frac{3}{2}$
(i) Determine the ratios $\frac{AD}{AB}and\frac{DE}{BC}.$
(ii) Prove that $\Delta DEF$ is similar to $\Delta CBF$ Hence, find $\frac{EF}{FB}$.
(iii) What is the ratio of the areas of $\Delta DFEand\Delta BFC$? [4 MARKS]

Answer 1

Application of theorems: 2 Marks
Calculation: 2 Marks
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Proof: In $\Delta ADE$ and $\Delta ABC$,
$\begin{array}{ccc}& \angle A=\angle A& \left[common\right]\\ & \angle 1=\angle 2& \left[Correspondingangles\right]\\ So,& \Delta ADE\sim \Delta ABC& \left[AAsimilarity\right]\\ Hence& \frac{AD}{AB}=\frac{DE}{BC}=\frac{AE}{AC}& \\ \left(i\right)Given& \frac{AD}{DB}=\frac{3}{2}& \\ Let& AD=3x.DB=2x& \\ So,& \frac{AD}{AB}=\frac{AD}{AD+DB}=\frac{3x}{3x+2x}=\frac{3x}{5x}=\frac{3}{5}& \\ Now,& \frac{AD}{AB}=\frac{DE}{BC}& \\ So,& \frac{DE}{BC}=\frac{3}{5}& \end{array}$

(ii) In $\Delta DEF$ and $\Delta CBF$,
$\begin{array}{ccc}& \angle 3=\angle 4& \left[Verticallyoppositeangles\right]\\ & \angle 5=\angle 6& \left[Alternateinteriorangles\right]\\ So,& \Delta DEF\sim \Delta CBF& \left[AAsimilarity\right]\\ Hence& \frac{DE}{BC}=\frac{EF}{BF}=\frac{DF}{CF}& \\ Now,& \frac{DE}{BC}=\frac{EF}{FB}& \\ So,& \frac{EF}{FB}=\frac{3}{5}& [\because \frac{DE}{BC}=\frac{3}{5}]\\ \left(iii\right)& \frac{ar\left(\Delta DEF\right)}{ar\left(\Delta BCF\right)}=\frac{D{E}^2}{B{C}^2}={\left(\frac{3}{5}\right)}^2=\frac{9}{25}& \end{array}$