# Geometry and triangles problem

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.

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