1. parallel lines

In the figure , JG bisects $\displaystyle \angle DGH$ , the line FG is parallel EK , and KJ is parallel to GH . Prove that EJ is parallel to FH

2. Originally Posted by thereddevils
In the figure , JG bisects $\displaystyle \angle DGH$ , the line FG is parallel EK , and KJ is parallel to GH . Prove that EJ is parallel to FH
you start with 2 pairs of similar triangles here:
DEK and DFG
DJK and DHG
not difficult to prove them similar to each other, right? there are some parallel lines....

Then, you can prove that DE : DF = DJ : DH (= DK : DG). together with common angle D, you can eventually prove the 3rd pair of similar triangles, DEJ and DFH. And since they are similar, angles DEJ = DFH, and finally...

3. Originally Posted by ukorov
you start with 2 pairs of similar triangles here:
DEK and DFG
DJK and DHG
not difficult to prove them similar to each other, right? there are some parallel lines....

Then, you can prove that DE : DF = DJ : DH (= DK : DG). together with common angle D, you can eventually prove the 3rd pair of similar triangles, DEJ and DFH. And since they are similar, angles DEJ = DFH, and finally...
Thanks Ukorov , i got it . But this is the continuatino of the question whihc i am unsure of :

Prove also that $\displaystyle \frac{DE}{EF}=\frac{DG}{GH}$

4. Originally Posted by thereddevils
Thanks Ukorov , i got it . But this is the continuatino of the question whihc i am unsure of :

Prove also that $\displaystyle \frac{DE}{EF}=\frac{DG}{GH}$
sorry I did not notice your reply...
First, angles DGJ = HGJ (given),
next, angles KJG = HGJ (alternate angles, KJ//GH)
so you have isosceles triangle KJG because angles KJG = DGJ
and hence KG = KJ

$\displaystyle \frac{DE}{EF} = \frac{DK}{KG} = \frac{DK}{KJ} = \frac{DG}{GH}$