Similar Triangles

Probably the toughest thing to do here is to find a pair of similar triangles that will be of benefit to you for this particular problem. Try your best to locate similar triangles. Use the given info about the parallel lines to help you. 

That is the part where I am stuck on because finding the similar triangles is so hard

I agree with you. Identifying the similar triangles is the toughest step. I had some trouble myself. 

Anyway, we know  \(\overline{AG}\parallel\overline{CH}\) from the given info. This means that corresponding angles are congruent. Therefore, \(\angle A\cong\angle HCD\) . Via the reflexive property of congruence, \(\angle HDA\cong\angle HDA\) . Can you think of two triangles that would be similar knowing this? We are hunting for two triangles that share a vertex at \(\angle HDA\) . This is the part that I initially overlooked. I hope this hint hel

We have  two triangles to consider

First  ...triangle  AGF

Since JE  is a segment drawn parallel to base AG.....we have the following relationship :

FE / JE   = FA / AG

1 / JE  = FA / AG

1/ JE  = 5 /AG

AG = 5  (JE )        (1)

Next...we have triangle AGD

Since  HC  is a segment drawn parallel to  base AG...we have the following relationship :

DC / HC  = DA / AG

1 / HC  =  3 / AG 

AG = 3 ( HC )       (2)

Equating (2) and (3)

5 ( JE )  = 3 ( HC )

HC / JE  = 5 / 3

Thank you everyone.. It helped a lot XSquaredFactor and CPhill, here is my thought process on this one after XSquaredFactor gave me a hint