Question

In the given figure, DE || BC and AD : DB = 5 : 4, then $\frac{areaof\Delta DOE}{areaof\Delta DCE}$ is

• A. ABC
• B. AEF
• C. BFD, CDE
• D. AFE, BFD, CDE

Answer 1

The correct option is C

BFD, CDE

Given, $\frac{AD}{DB}=\frac{5}{4}$

$\Rightarrow \frac{AD}{AD+DB}=\frac{5}{5+4}\Rightarrow \frac{AD}{AB}=\frac{5}{9}$
In $\Delta ADE$ and $\Delta ABC$,
$\angle ADE=\angle ABC$ [corresponding angles]
and $\angle A=\angle A$ [common]
$\therefore \Delta ADE\sim \Delta ABC$, [by AA similarity]
$\Rightarrow \frac{DE}{BC}=\frac{AD}{AB}=\frac{5}{9}\Rightarrow DE:BC=5:9$
Now, in $\Delta DOE$ and $\Delta COB$,
$\angle OED=\angle OBC$ [alternate angles]
$\angle DOE=\angle BOC$ [vertically opposite angles]
$\therefore \Delta DOE\sim \Delta COB$ [by AA similarity]
$\Rightarrow \frac{DO}{OC}=\frac{DE}{BC}=\frac{5}{9}\Rightarrow \frac{DO}{OD+OC}=\frac{5}{5+9}\Rightarrow \frac{DO}{DC}=\frac{5}{14}$
Now, draw $EN\perp CD$
$\therefore \frac{Areaof\Delta DOE}{Areaof\Delta DCE}=\frac{\frac{1}{2}\times DO\times EN}{\frac{1}{2}\times DC\times EN}=\frac{DO}{DC}=\frac{5}{14}$