If D be the mid-point of AB and if the internal bisectors of $\angle ADC$ and $\angle BDC$ meet $AC$ and $BC$ at H and I respectively. Prove that $HI \parallel AB$

My attempt:


It is obvious that:

$\angle CDH = \angle HDA = a$ (let)
The fact that angles on a straight line sum to $180$ and that $\angle CDI = \angle IDB$ gives: Then, $\angle CDI = \angle IDB = 90 - a$

Now, $\angle CDH + \angle CDI = a + (90 - a) = 90$. Thus, $\triangle HID$ is right-angled.

This is all what I could, and I don't understand what to do next.

Any hint is appreciated.

  • $\begingroup$ At some stage you will need to use D being the mid-point of AB $\endgroup$
    – Henry
    Nov 27 '14 at 14:41
  • 2
    $\begingroup$ Are you familiar with the Angle Bisector Theorem? $\endgroup$
    – Blue
    Nov 27 '14 at 14:46
  • $\begingroup$ No, I didn't know that. Thanks for it. $\endgroup$ Nov 27 '14 at 15:06

the fact that $DI$ bisects $\angle BDC$ gives ${CD \over DB} = {CI \over BI}.$ same way,${CD \over AD} = {CH \over AH}.$ these two and $AD = BD$ shows ${CI \over BI} = {CH \over AH}.$ now you can conclude that $HI$ is parallel to $AB.$

  • $\begingroup$ Yes, I can now, thanks! $\endgroup$ Nov 27 '14 at 15:09

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