Finding perimeter of a parallelogram

Question

In triangle $ABC$,$AB=AC=28$ and $BC=20$.Points $D,E,F$ are on sides $AB$,$BC$ & $CA$ ,respectively,such that $DE$ & $EF$ is parallel to $AC$ & $AB$,respectively .What is the perimeter of the parallelogram $ADEF$?

I tried to find out any relation between $AD$ & $FC$,but failed.One time I thought about the area related property but I failed to find anyway.Somebody please give me some hint to do this.Thank you

Answer 1

• Show that BDE and EFC are isoceles

• Use it to write $AD+DE+EF+AF=AD+BD+AF+FC=AB+BC=56$

Answer 2

$56$ of course! We can calculate it by the following way: $$AD+AF+FE+DE=2(AF+FE)=2AC=56.$$

Answer 3

since $DE$ and $AC$,or $EF$ and $FC$ are parallels so $\triangle ABC,\triangle BDE,\triangle EFC$ are isosceles triangles.$$\begin{cases} 28-x=y \\ 28-y=x \end{cases}\Rightarrow x+y=28$$ so perimetr is

$${ P }_{ ADEF }=2\left( x+y \right) =56$$