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# If D,E,F are the mid-points of sides BC,CA and AB respectivelyof $\Delta ABC$,then the ratio of the areas of triangles DEF and ABC is

$\begin{array}{1 1}1 : 4\\1 : 2\\2 : 3\\4 : 5\end{array}$

$FE=\large\frac{1}{2}$$BC$
=> $\large\frac{FD}{AC} =\large\frac{1}{2}$
$\large\frac{FE}{BC} = \frac{ED}{AB} =\frac{FD}{AC}$
$\qquad= \large\frac{1}{2}$
By similarly $ABC \approx \Delta FDE$