See the figure below:
Let $I_1F = r_1$, $I_2G = r_2$, and $I_1E = r_3$.
Note that $GD =r_2$, $DF = r_1$, and $I_2E = r_3$.
Note also that $\triangle I_1DI_2$ is a right triangle.
So using Pytagoras' Theorem in triangles $\triangle DGI_2$, $\triangle DFI_1$, and $\triangle I_1DI_2$ we get:
$$DI_1= r_1 \sqrt{2},$$
$$DI_2= r_2 \sqrt{2},$$
$$I_1I_2= \sqrt{2(r_1^2+r_2^2)} \tag1$$
Let $m(\angle I_1AI_2)= \alpha$, as $AI_2$ is bisector of $\angle CAD$ and $AI_1$ is bisector of $\angle BAD$, we can conclude that
$$\alpha = 45 ^{\circ}$$
But if $\alpha = 45 ^{\circ}$ then $\angle I_1EI_2$ is a right angle (central angle), and using Pytagoras' Theorem in $\triangle I_1EI_2$ and equation $(1)$ we get:
$$r_3=\sqrt{r_1^2+r_2^2} \tag2$$
Now let's calculate $r$, the inscribed circle radius of $\triangle ABC$, and compare it with $r_3$.
See the picture below:
We know that $\triangle ADC$, $\triangle BDA$, and $\triangle BAC$ are similar, then
$$\frac{r_1}{c}=\frac{r_2}{b}=\frac{r}{a}=k \tag3$$
So using relation $(3)$ and Pytagoras' Theorem again ($\triangle ABC$) we get:
$$a^2=b^2+c^2 \Rightarrow \left(\frac{r}{k}\right)^2=\left(\frac{r_1}{k}\right)^2 + \left(\frac{r_2}{k}\right)^2 \Rightarrow $$
$$\Rightarrow r^2=r_1^2+r_2^2 \Rightarrow r=\sqrt{r_1^2+r_2^2} \tag4$$
Therefore comparing $(2)$ and $(4)$ we can conclude finally that
$$r=r_3.$$
It's well known that $DE, IF, CM$ are concurrent. Call this point $X$.
Then $CX\perp IK$, $CI\perp XK$, meaning $I$ is the orthocenter of $CXK$ and the result follows.
Another interesting thing is that $\triangle CXK$ is self-polar wrt the incircle.
Edit:
The following proof that $DE,IF,CM$ are concurrent is not mine. It's taken from Evan Chen's EGMO.
Let $X$ be the intersection of $DE$ and $FI$ and let $A',B'$ be points on $CA$ and $CB$ such that $A'B'\parallel AB$ and $A'B'$ passes through $X$.
Now $E$ is the foot of $I$ onto $CA'$, $X$ is the foot of $I$ onto $A'B'$ ($B'A'\parallel BA$ and $IF\perp AB$, so $IF\perp AB$), and $D$ is the foot of $I$ onto $CB'$. Then $EXD$ is in fact the Simson line of $I$ wrt $\triangle CA'B'$, which means that $I$ is on the circumcircle of $\triangle CA'B'$.
Since $CI$ is an angle bisector of $\angle A'CB'$, $I$ is the arc midpoint of arc $B'A'$, and so $XI$ is the perpendicular bisector of $B'A'$. This means $X$ is the midpoint of $B'A'$, and so taking a homothety at $A$ which takes $A'B'\mapsto AB$ shows that the $C$-median passes through $X$ as well.
Once seeing $I$ is the orthocenter of $\triangle CXK$, $CK\perp IX$ but we already know $IX\perp AB$ so $CK\parallel AB$.
Best Answer
Let $D$ be a point on the line segment $AB$ such that $VD=\frac{VA+VB}{2}$.
(Such a point $D$ exists. The reason is as follows : Let $A'$ be a point on the half line $VA$ such that $VA'=\frac{VA+VB}{2}$. Let $B'$ be a point on the half line $VB$ such that $VB'=\frac{VA+VB}{2}$. Then, we have $VA'\gt VA$ since $$2VA\lt VA+VB\implies VA\lt \frac{VA+VB}{2}=VA'$$ Similarly, $VB'\lt VB$ since $$2VB\gt VA+VB\implies VB\gt\frac{VA+VB}{2}=VB'$$ Now on the plane $VAB$, let us consider a circle whose center is $V$ with radius $\frac{VA+VB}{2}$. We see that $A',B'$ are on the circle. Since $VA'\gt VA$ and $VB'\lt VB$, there is a point $D$ on the line segment $AB$ such that $D$ is on the circle. For such a point $D$, we have $VD=\frac{VA+VB}{2}$.)
Let $E$ be a point on the line segment $VD$ such that $VE=\frac 23VD$.
Let $F$ be a point on the line segment $VC$ such that $VF=\frac 13VC$.
Let $G$ be a point on the half line $VD$ such that $VG=VE+VF$.
Let $H$ be a point on the half line $VC$ such that $VH=VE+VF$.
Now, since $VD=\frac{VA+VB}{2}$, we have $$\frac{VA+VB+VC}{3}=\frac{2VD+VC}{3}=VE+VF$$
We have $VG\gt VD$ since $$\begin{align}&VC-VA+VC-VB\gt 0 \\&\implies VC\gt \frac{VA+VB}{2} \\&\implies VC\gt VD \\&\implies \frac 13VC\gt \frac 13VD \\&\implies VF\gt ED \\&\implies VF+VE\gt ED+VE \\&\implies VG\gt VD\end{align}$$
Also, we have $VC\gt VH$ since $$\begin{align}\frac 23VC\gt\frac 23VD&\implies FC\gt VE \\&\implies FC+VF\gt VE+VF \\&\implies VC\gt VH\end{align}$$
Now, on the plane $VCD$, let us consider a circle whose center is $V$ with radius $VE+VF$.
$G,H$ are on the circle with $VG\gt VD$ and $VH\lt VC$.
So, there is a point $P$ on the line segment $CD$ such that $P$ is on the circle.
For such a point $P$, we have $$VP=VE+VF=\frac{VA+VB+VC}{3}. \blacksquare$$