Proof. Note that the incenter, $I$ is determined by the intersection of the internal angle bisectors of the triangle. We will consider the $A$-excircle, i.e. the excircle whose center, let's call it $X$, is determined by the intersection of the external angle bisectors at vertices $B$ and $C$. Note that $X$ lies on the line $AI$ as well.
Let $\angle BAC = \alpha$, $\angle ABC = \beta$ and $\angle ACB = \gamma$. Furthermore, let $AX \cap (ABC) = M$, so we need to show that $MI = MX$. We claim that $BICX$ is a cyclic quadrilateral centered at $M$. Firstly, we have $\angle BAM = \angle CAM = \frac{\alpha}{2}$, so $MB = MC$. Next, we show that $\triangle MCI$ is isosceles. Indeed, we have $\angle MIC = \angle IAC + \angle ACI = \frac{\alpha}{2} + \frac{\gamma}{2} = \angle BAM + \angle BCI = \angle BCM + \angle BCI = \angle MCI$.
Since $MB = MC = MI$, $M$ is the circumcenter of $\triangle BCI$, it remains to show that $X ∈ (BCI)$. We have $\angle BIC = 180° - \frac{\beta}{2} - \frac{\gamma}{2}$. On the other hand, $\angle BXC = 180° - \angle CBX - \angle BCX = 180° - \frac{180° - \beta}{2} - \frac{180° - \gamma}{2} = \frac{\beta}{2} + \frac{\gamma}{2}$.
Therefore, $\angle BIC + \angle BXC = 180°$, i.e. $X ∈ (BCI)$ and $(BCI)$ is centered at $M$, so $MI = MX$, as desired.
This answer uses $X,Y,Z$ that you defined.
Let $E$ be a point on $BC$ such that $ME\parallel CD$.
Let $F$ be a point on $CD$ such that $MF\parallel BD$.
Let $G$ be a point on $BD$ such that $MG\parallel BC$.
Then, we have $$AD:AX=BC:BE\tag1$$ (Proof : Let us consider the plane $\alpha$ on which $A,A',B$ and $M$ exist. Let $P$ be the intersection point of $\alpha$ with $CD$. Then, we have $PA:AA'=PB:MB$. We also have $PA:AA'=AD:AX$ and $PB:MB=BC:BE$. So, we get $AD:AX=BC:BE$.$\ \square$)
Similarly, we have
$$AD:AX=CD:CF\tag2$$
$$AC:AY=DB:DG\tag3$$
Now, if $(BCD)\parallel (A'B'C')$, then we can say that $AD:AX=AC:AY$.
So, it follows from $(1)(2)(3)$ that
$$BC:BE=CD:CF=DB:DG\tag4$$
Let $H$ be a point on $BD$ such that $MH\parallel CD$.
Let $I$ be a point on $BC$ such that $MI\parallel BD$.
Let $J$ be a point on $CD$ such that $MJ\parallel BC$.
We have $$CJ=FD\tag5$$
By Menelaus's theorem, we get
$$\frac{CF}{FP}\times\frac{PM}{MB}\times\frac{BI}{IC}=1$$
Since $\frac{BI}{IC}=\frac{DF}{CF}$, we have
$$\frac{PM}{MB}=\frac{FP}{DF}\tag6$$
By Menelaus's theorem, we get
$$\frac{DJ}{JP}\times\frac{PM}{MB}\times\frac{BG}{GD}=1$$
Since $\frac{BG}{GD}=\frac{CJ}{JD}$, we have
$$\frac{PM}{MB}=\frac{JP}{CJ}\tag7$$
It follows from $(5)(6)(7)$ that
$$JP=FP\tag8$$
It follows from $(5)(8)$ that $P$ is the midpoint of $CD$.
Similarly, we can see that the intersection point of $CM$ with $BD$ is the midpoint of $BD$.
Therefore, we can say that $M$ is the centroid of the triangle $BCD$.
Added :
how do you know in the first proof that $P$ is collinear with $A,A'$ and $B,M$? How do you know that the intersection of $AA'$ with $BM$ is point $P$ which is on $CD$?
Since $A'M\parallel AB$, the line $AA'$ intersects the line $BM$. Let $Q$ be the intersection point of $AA'$ with $BM$. We see that $Q$ is on $\alpha$. We also see that $Q$ is both on the plane $ACD$ and on the plane $BCD$. So, we see that $Q$ is on the line $CD$. It follows that $P=Q$. So, $P$ is both on the line $AA'$ and on the line $BM$.
I read what you wrote forward and saw that you dont think that $BM$ interesects $AA'$ in point $P$
I replaced $K$ with $P$ since $K$ is nothing but $P$.
Can you please explain how you got the following ratios $PA:AA′=PB:MB, PA:AA′=AD:AX$ and $PB:MB=BC:BE$
$PA:AA′=PB:MB$ since $\triangle{PAB}\sim\triangle{PA'M}$.
$PA:AA′=AD:AX$ since $\triangle{AA'X}\sim\triangle{APD}$.
$PB:MB=BC:BE$ since $\triangle{BME}\sim\triangle{BPC}$.
Best Answer
First we prove $OG\perp A'B'$.
We have $\overrightarrow{OG}=\frac{1}{3}(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})$,
$\overrightarrow{OG}\cdot\overrightarrow{A'B'}=\frac{1}{3}(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})\cdot(\overrightarrow{A'O}+\overrightarrow{OB'})=\frac{1}{3}(\overrightarrow{OA}\cdot\overrightarrow{A'O}+\overrightarrow{OB}\cdot\overrightarrow{OB'})=\frac{1}{3}(\overrightarrow{OB}\cdot\overrightarrow{OB'}-\overrightarrow{OA}\cdot\overrightarrow{OA'})=0$
from $OA, OB, OC$ are perpendicular to each other and $A,B,B',A'$ are on the same circle so $\triangle OAB\sim\triangle OB'A'$.
Similarly, $OG\perp B'C'$ and $OG\perp A'C'$.
Let's denote $H'$ as the intersection of $OG$ and $(A'B'C')$, and prove that $H'$ is the orthocenter of $\triangle A'B'C'$. Now that we have $OG\perp (A'B'C')$, $H'$ is the projection of $O$ onto plain $(A'B'C')$, so to prove $A'H\perp B'C'$, we just need to prove that $OA'\perp B'C'$.
$\overrightarrow{OA'}\cdot\overrightarrow{B'C'}=\overrightarrow{OA'}\cdot(\overrightarrow{B'O}+\overrightarrow{OC'})=\overrightarrow{OA'}\cdot\overrightarrow{B'O}+\overrightarrow{OA'}\cdot\overrightarrow{OC'}=0$.
Similarly we can prove $B'H\perp A'C'$, $C'H\perp A'B'$. The proof that $H'$ is the orthocenter of $\triangle A'B'C'$ is complete, thus $H$ is on $OG$.
(Without using vectors:)
To prove $OG\perp B'C'$, denote $D$ as the midpoint of $BC$. $OB\perp OC$, so $OD=DB=DC$, $\angle BOD=\angle OBD=\angle OC'B'$, thus $OD\perp B'C'$. Since $OA\perp (OBC)$, $OA\perp B'C'$, thus $(OAD)\perp B'C'$, $OG\perp B'C'$. The rest goes similarly.
For the next part, just notice that $OA'\perp (OB'C')$ and $H'$ is the projection of $O$ onto $(A'B'C')$, we must have $AH'\perp B'C'$. The rest goes similarly.