Question

If D, E and F are respectively the midpoints of sides AB, BC and CA of △ABC then what is the ratio of the areas of △DEF and △ABC?

Answer 1

By using mid theorem i.e., the segment joining two sides of a triangle at the midpoints of those sides is parallel to the third side and is half the length of the third side.
$$\begin{gathered} \therefore \mathrm{DF}\parallel \mathrm{BC} \\ \mathrm{and} \\ \mathrm{DF}=\frac{1}{2}\mathrm{BC} \\ \Rightarrow \mathrm{DF}=\mathrm{BE} \end{gathered}$$
Since, the opposite sides of the quadrilateral are parallel and equal.
Hence, BDFE is a parallelogram
Similarly, DFCE is a parallelogram.
Now, In △ABC and △EFD
∠ABC = ∠EFD (Opposite angles of a parallelogram)
∠BCA = ∠EDF (Opposite angles of a parallelogram)
By AA similarity criterion, △ABC ∼ △EFD
If two triangles are similar, then the ratio of their areas is equal to the ratio of the squares of their corresponding sides.
$\therefore \frac{\mathrm{area}\left(△\mathrm{DEF}\right)}{\mathrm{area}\left(△\mathrm{ABC}\right)}={\left(\frac{\mathrm{DF}}{\mathrm{BC}}\right)}^2={\left(\frac{\mathrm{DF}}{2\mathrm{DF}}\right)}^2=\frac{1}{4}$
Hence, the ratio of the areas of △DEF and △ABC is 1 : 4.