Given: df and de are midsegments of abc prove: de=1/2 ac

Solution-

Given, DF and DE are midsegments of ABC. so D is the mid point of AB and F is the mid point of AC.

Considering AB,

Then DE is parallel to AC, as mid-segments are parallel to the non included side of the triangle.

Also, according to corresponding angle theorem,

Then,

As,

According to similarity of two triangles,

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x