Question : D, E, and F are the midpoints of the sides BC, CA, and AB, respectively of a $\triangle ABC$. Then the ratio of the areas of $\triangle DEF$ and $\triangle ABC$ is:
Option 1: $\frac{1}{2}$
Option 2: $\frac{1}{4}$
Option 3: $\frac{1}{8}$
Option 4: $\frac{1}{16}$
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Correct Answer: $\frac{1}{4}$
Solution : Given: D, E, and F are the midpoints of the sides BC, CA, and AB, respectively of a $\triangle ABC$.
$DF || BE$ and $DF = \frac{1}{2}BC$
Since the opposite sides of the quadrilateral are equal and parallel.
So, BDFE is a parallelogram.
Similarly, DFCE is also a parallelogram.
In $\triangle ABC$ and $\triangle EFD$,
$\angle ABC= \angle EFD$ (opposite angles of a parallelogram)
$\angle BCA= \angle EDF$ (opposite angles of a parallelogram)
$\angle BAC= \angle DEF$ (opposite angles of a parallelogram)
So, $\triangle ABC\sim \triangle EFD$ (by AAA similarity)
The ratio of the squares of the corresponding sides of two similar triangles equals the ratio of their areas.
⇒ $\frac{\text{area of $\triangle DEF$}}{\text{area of $\triangle ABC$}}=\frac{(DF)^2}{(BC)^2}$
$=\frac{(DF)^2}{(2DF)^2}$
$=\frac{1}{4}$
The ratio of the areas of $\triangle DEF$ and $\triangle ABC$ is 1 : 4.
Hence, the correct answer is $\frac{1}{4}$.
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