Problem in similarity of triangles

[agnibho]

1. The problem statement, all variables and given/known data
ABCD is a parallelogram. E, F are points on the straight line parallel to AB. AF, BF meet at P, and DE, CF meet at Q. Prove that PQ ll AD.

2. The attempt at a solution
I drew the diagram. 36154
I tried to solve the problem in this way:-
CD ll XY Therefore, angleCDE = angleDEF (alternate interior angles)
Also, angleDCF = angleCFE (alt. int. angles)

Hence, triangle DQC is similar to triangle FQE
So, DQ/QE = CQ/QF

After this I felt at a loss. I couldn't figure out the next step. I thought that if I could anyhow prove that the angleEQP = angleEDA, then I could've said that they are equal but they are corresponding angles and hence I could've proved PQ ll AD.

Someone please help me with the next step .

[tiny-tim]

hi agnibho! :smile:

(btw, that's a bad diagram … you should have drawn it so that DE is obviously not parallel to BF :wink:)
CD ll XY Therefore, angleCDE = angleDEF (alternate interior angles)
Also, angleDCF = angleCFE (alt. int. angles)

Hence, triangle DQC is similar to triangle FQE
So, DQ/QE = CQ/QF

hint: you haven't yet used any connection between the top and bottom halves of the diagram :wink:

[agnibho]

All right....I will try that....and about the parallel issue I think I really did a mistake there!!:biggrin:

[agnibho]

But what about the hint?? Did you mean to say that I can use the alternate angles theorem again for the connection too??:confused:

[tiny-tim]

i was thinking about the length of EF (which is shared between the top and bottom halves of the figure) wink: