Question:
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.
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 $AB.$