How to prove that three points are collinear? I have this homework problem where the details are in the attached diagram.

How do you prove that the 2 orthocenters are tangent to the tangent point of the incircle to the side BD?
 A: I will perform a little change of notation. We have a triangle $ABC$ with incenter $I$. $I_{A}$ is the tangency point of the incircle on the $BC$ side, $H_C$ is the orthocenter of $ABI$ and $H_B$ is the orthocenter of $ACI$. We want to prove that $I_A,H_B,H_C$ are collinear.

Both $BH_C$ and $CH_B$ are $\perp IA$. We have $AH_C\perp IB$ and $AH_B\perp IC$, with $\widehat{BIC}=\frac{\pi}{2}+\widehat{BAC}$.
If we prove that the ratio $\frac{BH_C}{CH_B}$ equals the ratio $\frac{BI_A}{CI_A}=\frac{a+c-b}{a+b-c}$ we are done by similarity.
On the other hand, by angle chasing we have that the symmetric of the orthocenter of a triangle with respect to a side lies on the circumcircle. So it is worth to consider the circumcircles of $AIB$ and $AIC$ and $K_B, K_C$ as in the following diagram:

We have $CI_A=CI_B, BI_A=BI_C$ and by symmetry $CH_B=CK_B$ and $BH_C=BK_C$. It is enough to show that $CI_BK_B$ and $BI_C K_C$ are similar, and that simply follows from
$$\widehat{CK_B I_B}=\widehat{CAI}=\widehat{BAI}=\widehat{BK_C I_C}.$$
A: Some notations: Let the incircle of triangle $ABD$ touches the edges $AB, \, AD$ and $BD$ at the points $P, \, Q$ and $T$ respectively. Thus, $CP$ is the altitude of triangle $ABC$ through vertex $C$ and $CQ$ is the altitude of triangle $ACD$ through vertex $C$. The altitude of triangle $ABC$ through vertex $B$ intersects the other altitude $CP$ at point $H_{ABC}$, which is the orthocenter of $ABC$. The altitude of triangle $ACD$ through vertex $D$ intersects the other altitude $CQ$ at point $H_{ACD}$, which is the orthocenter of $ACD$. 

Proof: Let us look at the two triangles $BPH_{ABC}$ and $DQH_{ACD}$. Clearly, $$\angle \, BPH_{ABC} = 90^{\circ} = \angle \, DQH_{ACD}$$
Since $AC$ is the angle bisector of $\angle \, BAD = 2 \gamma$, 
$$\angle \, DAC = \gamma = \angle \, BAC$$
we can conclude that 
$$\angle \, PBH_{ABC} = \angle \, ABH_{ABC} = 90^{\circ} - \gamma = \angle \,  ADH_{ACD}= \angle \, QDH_{ACD}$$
Thus triangles  $BPH_{ABC}$ and $DQH_{ACD}$ are similar and so
$$\frac{BH_{ABC}}{DH_{ACD}} = \frac{BP}{DQ}$$
However, since  $BP = BT$ and $DQ = DT$, because $BP$ and $BT$ are the two tangents to the incircle through point $B$, and because $DQ$ and $DT$ are the two tangents to the incircle through point $D$.
Consequently,
$$\frac{BH_{ABC}}{DH_{ACD}} = \frac{BP}{DQ} =  \frac{BT}{DT}$$ which by Tales' intercept theorem is possible if and only if the three points $N_{ABC}, \, T$ and $H_{ACD}$ are collinear.
