^@ D^@, ^@E^@ and ^@F^@ are respectively the midpoints of sides | Math Question and Answer

Answer:

$1:4$

Step by Step Explanation:

In $\Delta ABC$ , $D$ and $F$ are the midpoints of sides $AB$ and $CA$ respectively.

Therefore, $DF || BC$ [ By midpoint theorem ]

$\implies\space\space DF || BE.$

Similarly, $\space\space EF || BD$ .

Therefore, $BEFD$ is a parallelogram.

$\implies \angle B = \angle EFD, EF = BD = \dfrac { 1 } { 2 } AB$ and $DF = BE = \dfrac { 1 } { 2 } BC$ .

Also, $ECFD$ is a parallelogram.

$\implies\space\space \angle EDF = \angle C.$
Now, in $\Delta DEF$ and $\Delta CAB$ , we have $\begin{aligned} &\angle EFD = \angle B \\ and \space\space& \angle EDF = \angle C \\ \therefore \space\space& \Delta DEF \sim \Delta CAB &&[\text{ By AA-similarity}] \\ \text{ So, }& \dfrac { ar(\Delta DEF ) } { ar( \Delta ABC ) } = \dfrac {ar( \Delta DEF )} { ar(\Delta CAB) } = \dfrac { DF^2 } { BC^2 } = \dfrac{ (\dfrac{1} { 2 } BC)^2 } { BC^2 } = \dfrac { 1 } { 4 } \end{aligned}$ Thus, the ratio of the areas of $\Delta DEF$ and $\Delta ABC$ is $1:4$ .

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