Question : $D$ and $E$ are points on the sides $AB$ and $AC$ respectively of $\triangle ABC$ such that $DE$ is parallel to $BC$ and $AD: DB = 4:5$, $CD$ and $BE$ intersect each other at $F$. Find the ratio of the areas of $\triangle DEF$ and $\triangle CBF$.
Option 1: $16:25$
Option 2: $16:81$
Option 3: $81:16$
Option 4: $4:9$
Correct Answer: $16:81$
Solution :
In $\triangle ADE$ and $\triangle ABC$
$\angle AED=\angle C$ (corresponding angle)
$\angle ADE=\angle B$ (corresponding angle)
$\angle A=\angle A$ (common angle)
$\triangle ADE\sim\triangle ABC$ (by AAA criteria)
$\frac{AD}{AB}=\frac{AE}{AC}=\frac{DE}{BC}=\frac{4}{9}$
In $\triangle DEF$ and $\triangle BFC$
$\angle EDF=\angle BCF$ (alternate interior angle)
$\angle DEF=\angle CBF$ (alternate interior angle)
$\triangle DEF\sim\triangle CBF$ (by AA criteria)
The theorem used is the two triangles are similar, and the ratio of areas is equal to the ratio of squares of corresponding sides.
$\frac{ar\triangle DEF}{ar\triangle BFC}=(\frac{DE}{BC})^2=(\frac{4}{9})^2=\frac{16}{81}$
Hence, the correct answer is $16:81$.
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