Let's Look Into The Incircle

Let us look into some of the properties of the incircle and excircles of a triangle.

Lemma $1$ : Let $ABC$ be a triangle with incenter $I$. The incircle of $\triangle ABC $ is tangent to $ BC, CA $ and $ AB $ at $D,E$ and $F$ respectively. Let $ M_a , M_b $ and $M_c $ be midpoints of $ \overline{ \rm BC} , \overline{ \rm CA} $ and $ \overline{ \rm AB }$ respectively. Let ray $ BI $ and $CI $ meets line $EF$ at $X$ and $Y$ respectively. Then we have :

1. $BX \perp CX $ and $ CY \perp BY $.
2. $X$ lies on $M_aM_b $ and $Y$ lies on $ M_aM_c $.
3. $I$ is incenter of $ \triangle DXY $.

Proof : 

To avoid configuration issue  we are going to use directed angles.

Claim $1$ : $ I, E, X, C $ are concyclic.

Proof : 

$ \measuredangle IXE = \measuredangle BXF = -\measuredangle XFB - \measuredangle FBX = 90^{\circ} -\frac{1}{2} \angle A -\frac{1}{2} \angle B = \frac{1}{2} \angle C = \measuredangle ICE  $

Hence $ I,E,X,C $ are concyclic.

$ \therefore \measuredangle BXC = \measuredangle IXC = \measuredangle IEC = 90^{\circ} $

$ \therefore BX \perp CX $

Similarly $ CY \perp BY $, which completes part $1$ of our lemma.

Now $ \measuredangle BXC = 90^{\circ} \Rightarrow X \in $ circle with radius $ = BM_a = CM_a $ and centered at $ M_a $.

$ \therefore M_aX = BX \Rightarrow \angle M_aXB = \angle XBM_a  = \angle XBF $

$ \therefore AB \parallel M_aX \Rightarrow M_b \in M_aX $

Hence $ M_a , M_b , X $ are collinear.

Similarly $ M_a , M_c , Y $ are collinear, which completes the proof of part $2$.

Now $ \measuredangle IXC = 90^{\circ} = \measuredangle IDC $

Hence $IXCD $ is a cyclic quadrilateral.

$ \therefore \measuredangle DXI = \measuredangle DCI= \frac{1}{2} \angle C = \measuredangle IXY $

Hence $ IX $ bisects $ \angle DXY $, similarly $ IY $ bisects $ \angle DYX $, thus $I$ is incenter of $ \triangle DXY $, which completes our lemma.

Lemma $2$ : ( Duality of Orthocenters and Excenters) Let $I_a , I_b , I_c $ be the excenters of $\triangle ABC $, then triangle $ABC$ is orthic triangle of $ \triangle I_aI_bI_c $, and the orthocenter is $I$, where $I$ is incenter of triangle $ABC$.

Proof :

The proof is very simple, observe that $ A, I $ and $I_a$ are collinear.

Also observe that $ I_b , A , I_c$ are collinear.

And $ \angle I_a A I_b = \angle I_a A C + \angle CAI_b = 90^{\circ} $.

Hence $ I_aA \perp I_bI_c $. Similarly we get $ I_bB \perp I_cI_a $ and $ I_cC \perp I_aI_b $

Thus $ \triangle ABC $ is orthic triangle of $\triangle I_aI_bI_c $ and the orthocenter is $I$.

Lemma $3$ : Let $ABC$ be a triangle whose incircle is tangent to $ \overline{ \rm BC} $ at $D$. If $ \overline{ \rm DE } $ is diameter of the incircle and ray $AE $ meets $BC$ at $X$, then $ BD = CX $ and $X$ is the tangency point of the $A$-excircle to $BC$. Also if $ XY $ is diameter of $A$-excircle of triangle $ABC$, then $ A,D,Y $ are collinear.

Proof :

We are going to use homothety to solve the problem.

Let $ B'C' \parallel BC $ and $ E \in B'C' $.

Then we have $ B'C' $ is tangent to incircle of triangle $ABC$, moreover the incircle of $\triangle ABC $ is $A$-excircle of $ \triangle AB'C' $ .

Now consider the homothety at $A$ mapping $ B'C' \rightarrow BC $.

Then $ E $ is mapped to extouch of $A$-excircle of  $\triangle ABC $ on $BC$, which is precisely $X$.

Hence $ A,E,X $ are collinear.

Now as $D$ and $X$ are isotomic conjugates of $\triangle ABC \Rightarrow BD = CX $.

Now let $ XY $ be the diameter of  $A$-excircle of $ \triangle ABC $. Then the homothety at $A$ mapping the incenter of $\triangle ABC $ to $ A$-excircle of $\triangle ABC $, maps $ E \rightarrow Y $.

As $ DE $ is diameter of the incircle and $ XY$ is diameter of the excircle, we deduce that the following homothety maps $ D \rightarrow Y $.

Hence $ A, D $ and $Y$ are collinear.

Hence, proved.

Lemma $4$ : Let $ABC$ be a triangle with incenter $I$ and $A$-excenter $I_a$, let $ D$ and $X$ be the associated tangency points on $BC$. Then lines $DI_a $ and $XI$ concur at the midpoint of altitude from $A$.

Proof :

So we are again going to use homothety.

Let $K$ be the foot of perpendicular from $A$ to $BC$ and $M$ be midpoint of $AK$.

We have $DE \parallel AK$ and  $ X,D,K $ are collinear and $ X,E,A $ are collinear ( by Lemma 3 ).

Hence consider the homothety at $X$ mapping $ DE \rightarrow AK $, then this homothety maps $I$ to midpoint of $AK$, which is $M$.

Hence $X,I,M $ are collinear.

Also we have $ AK \parallel XY $ and $ X,D,K $ are collinear and $ A,D,Y $ are collinear ( by Lemma 3 ).

Hence consider the homothety at $D$ mapping $ XY \rightarrow AK $.

Then $I_a$ is mapped to midpoint of $AK$, which is $M$.

Hence $ D,I_a,M $ are collinear.

Thus $ DI_a \cap XI = M $.

Hence, proved.