Question

D, E, F are respectively the mid-point of the sides AB, BC and CA of \[\Delta {\text{ABC}}\]. Find the ratios of the area of \[\Delta {\text{DEF}}\] and \[\Delta {\text{ABC}}\].

Answer 1

Hint:
First, we need to find the ratio of area of \[\Delta {\text{DEF}}\] and \[\Delta {\text{ABC}}\], so we first we need to prove these triangles are similar. We will use that the line joining midpoints of two sides of a triangle is parallel to the ${3}^{rd}$ side and that in a parallelogram, opposite angles are equal and we will then prove that \[\Delta {\text{DEF}}\] and \[\Delta {\text{ABC}}\] are similar using the AA criterion. Then we will use that if two triangles are similar, the ratio of their area is always equal to the square of the ratio of their corresponding side to find the required ratio.

Complete step by step solution:
We are given that D, E, F are respectively the mid-point of the sides AB, BC and CA of \[\Delta {\text{ABC}}\].
Since we need to find the ratio of area of \[\Delta {\text{DEF}}\] and \[\Delta {\text{ABC}}\], so we first we need to prove these triangles are similar.
We know that line joining midpoints of two sides of a triangle is parallel to the ${3}^{rd}$ side.
In \[\Delta {\text{ABC}}\],
We have D and F are midpoints of AB and AC, so the line DF is parallel to BC and DF is also parallel to BE.
Similarly, we have E and F are midpoints of BC and AC, so the line EF is parallel to AB and EF is also parallel to DB.
Thus, we have found out that the opposite sides of the quadrilateral are parallel.
Therefore, DBEF is a parallelogram.
Now we know that in a parallelogram, opposite angles are equal.
Hence, \[\angle DFE = \angle ABC{\text{ .......(1)}}\].
Similarly, we can prove that DECF is a parallelogram.
We know that in a parallelogram, opposite angles are equal.
Hence, \[\angle EDF = \angle ACB{\text{ .......(2)}}\].
Now, in \[\Delta {\text{DEF}}\] and \[\Delta {\text{ABC}}\],
\[\angle DFE = \angle ABC\]
\[\angle EDF = \angle ACB\]
By using AA similarity, we have that \[\Delta DEF \sim \Delta ABC\].
We know that if two triangles are similar, the ratio of their area is always equal to the square of the ratio of their corresponding side.
\[ \Rightarrow \dfrac{{Area\left( {\Delta DEF} \right)}}{{Area\left( {\Delta ABC} \right)}} = \dfrac{{D{E^2}}}{{A{C^2}}}\]
Since we have that DECF is a parallelogram, opposite sides are equal, we get
\[ \Rightarrow \dfrac{{Area\left( {\Delta DEF} \right)}}{{Area\left( {\Delta ABC} \right)}} = \dfrac{{F{C^2}}}{{A{C^2}}}\]
As we know that F is the midpoint of AC, we get
\[
   \Rightarrow \dfrac{{Area\left( {\Delta DEF} \right)}}{{Area\left( {\Delta ABC} \right)}} = \dfrac{{{{\left( {\dfrac{{AC}}{2}} \right)}^2}}}{{A{C^2}}} \\
   \Rightarrow \dfrac{{Area\left( {\Delta DEF} \right)}}{{Area\left( {\Delta ABC} \right)}} = \dfrac{{{{\dfrac{{AC}}{4}}^2}}}{{A{C^2}}} \\
   \Rightarrow \dfrac{{Area\left( {\Delta DEF} \right)}}{{Area\left( {\Delta ABC} \right)}} = \dfrac{1}{4} \\
 \]

Hence, the required ratio is \[1:4\].

Note:
We need to know that the AA criterion states that if two triangles have two pair of congruent angles, then the triangles are similar. We know that the similarity of triangles is different from the congruence of triangles. Here, the triangles are partially similar but congruent, the triangles are exact same. Be careful while writing the values and using the properties.