In Δ𝐴𝐵C the mid points of of sides 𝐴𝐵, 𝐵𝐶, and CA respectively. Find the ratio of $\mathrm{ar}\left(\mathrm{△}DEF\right)\text{ to }\mathrm{ar}\left(\mathrm{△}ABC\right)$.

𝐷, 𝐸, and F are respectively the midpoints of sides 𝐴𝐵, 𝐵𝐶, and CA of Δ𝐴𝐵𝐶

We know that, the line segment joining the midpoints of any two sides of a triangle is half the third side and parallel to it.

Therefore $FD=\frac{1}{2}BC,ED=\frac{1}{2}AC,EF=\frac{1}{2}AB$

In △𝐴𝐵𝐶 and △𝐸𝐹D

$\text{ We have }\frac{AB}{EF}=\frac{BC}{FD}=\frac{AC}{ED}=2$

△𝐴𝐵𝐶 and △𝐸𝐹𝐷 are similar by SSS congruence 

Also, We know that If two triangles are similar, then the ratio of the area of both triangles is equal to the square of the ratio of their corresponding sides.

$\frac{\mathrm{ar}\left(\mathrm{△}ABC\right)}{\mathrm{ar}\left(\mathrm{△}DEF\right)}={\left(\frac{AB}{EF}\right)}^2$

$\begin{array}{r}\Rightarrow \frac{\mathrm{ar}\left(\mathrm{△}ABC\right)}{\mathrm{ar}\left(\mathrm{△}DEF\right)}=4\\ \Rightarrow \frac{\mathrm{ar}\left(\mathrm{△}ABC\right)}{\mathrm{ar}\left(\mathrm{△}DEF\right)}=\frac{1}{4}\end{array}$

Hence, the ratio of 𝑎𝑟(△𝐷𝐸𝐹) to 𝑎𝑟(△𝐴𝐵𝐶) is 1:4