D,E and F are respectively the …

Given:D,E and F are respectively the mid-points of sides AB, AC and BC of ​{tex}\triangle {/tex}​ ABC.
To determine. Ratio of the area of ​{tex}\triangle {/tex}​ DEF and ​{tex}\triangle {/tex}​ ABC.
Determination:
We have
{tex}\frac{{AD}}{{AB}} = \frac{1}{2}{/tex}  ....(1)
{tex}\because {/tex} D is the mid-point of AB and
{tex}\frac{{AE}}{{AC}} = \frac{1}{2}{/tex} ....(2)

{tex}\because {/tex} E is the mid-point of AC
From (1) and (2),
{tex}\frac{{AD}}{{AB}} = \frac{{AE}}{{AC}}{/tex}
{tex}\therefore {/tex} DE {tex}\parallel{/tex} BC..........By converse of basic proportionality theorem
{tex}\therefore {/tex} {tex}\angle{/tex} ADE = {tex}\angle{/tex} ABC ....(3) .....corresponding angles
and {tex}\angle{/tex} AED= {tex}\angle{/tex} ACB  ........(4) ......corresponding angles
In view of (3) and (4),
​{tex}\triangle {/tex}​ ABC {tex} \sim {/tex} ​{tex}\triangle {/tex}​ DEF  ......AA similarity criterion
{tex}\frac{{ar(\triangle DEF)}}{{ar\left( {\triangle ABC} \right)}} = {\left( {\frac{{DE}}{{BC}}} \right)^2}{/tex}...... {tex}\because {/tex} The ratio of the areas of two similar triangle is equal to the square of the ratio of their corresponding sides.
{tex}= {\left( {\frac{{\frac{1}{2}BC}}{{BC}}} \right)^2}{/tex}................ {tex}\because {/tex} D and E are the mid-points of AB and AC respectively
  {tex}\therefore {/tex} DE||BC and DE={tex}\frac{1}{2}BC = \frac{1}{4}{/tex} {tex}\therefore {/tex} ar( ​{tex}\triangle {/tex}​ DEF):ar( ​{tex}\triangle {/tex}​ ABC)=1:4.