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,

i think it will be zero as slope has a formula of difference between initial and final distance divided by itz time interval