Thread: Prove similar triangles in parallelogram

Given a parallelogram ABCD and 2 segments FH and EG intersecting the diagonal AC at point P and terminating on 2 opposite parallelogram sides respectively, plus segments HG and EF.

Pprove that triangles EFP and HGP are similar.

Could I get a hint on how to solve this problem?

If it could be proven that HG and EF are parallel or by the same token, that angle AFE=GHC, the proof would follow. I have been unable to make any headway.

first, show using angle-angle.

showing these two triangles are similar will show

since they are alternate-interior angles

... now you can show . All you need is one more pair of angles (which should be obvious from the diagram)

Thank you skeeter. I still don't see how triangles HCG and FAE are proven similar. Angle EAF=HCG. But to say that angle AEF=CFH or AFE=CHG, would assume that EF is parallel to HG which I still don't see how to prove.

... you are correct. I'll have to stare at it awhile and see if I can draw a conclusion.

Thanks. I have been at it several days. I drew it in a computer and the lines (EF and FG) are indeed parallel, just can't figure out how to prove it.

Triangles AEP and CGP are similar so mAP = PC, mEP = GP and mAE = CG for some constant m.
Triangles APF and CPH are similar so nAP = PC, nAF = CH and nFP = HP for some constant n.
Equating PC from both sets of equations shows that m = n, in which case, (replacing n by m), mEP = GP and mFP = HP.
Now feed that information into the triangles EPF and GPH together with the common angles EPF and GPH and we can show that the two triangles are similar. (There might be some well known theorem that, for the moment, I've forgotten, but failing that use of the cosine rule in the two triangles shows that mEF = HG).

Thank you very much, BobP

Originally Posted by BERMES39

Thank you very much, BobP

one simple construction consisting of two lines makes the solution easy

I was wrong in my assumption about construction lines.Using Bob's similar triangles
Triangles AEP-CGP AP/PC =PF/HP
Triangles APF-CPH AP/PC= EP/PG
PF/HP=EF/PG
mEPF =mHPG
Triangles EPF -HPG
SAS theorem
If two sides and the included angle of one are in same proportion as the correspondingsides and included angle ofother they are similar