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Math Help - parallel lines

  1. #1
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    parallel lines

    In the figure , JG bisects \angle DGH , the line FG is parallel EK , and KJ is parallel to GH . Prove that EJ is parallel to FH
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    Quote Originally Posted by thereddevils View Post
    In the figure , JG bisects \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...
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    Quote Originally Posted by ukorov View Post
    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 \frac{DE}{EF}=\frac{DG}{GH}
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    Quote Originally Posted by thereddevils View Post
    Thanks Ukorov , i got it . But this is the continuatino of the question whihc i am unsure of :

    Prove also that \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

    \frac{DE}{EF} = \frac{DK}{KG} = \frac{DK}{KJ} = \frac{DG}{GH}
    Last edited by ukorov; November 21st 2009 at 10:56 AM.
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