It is well known that the symmetrical points $H_{1}, H_{2}, H_{3}$ of $H$ with respect to the sides $B C, A B, A C$ of the triangle $A B C$ respectively lie on the circle (c).
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Figure 8: G8
Let $L$ be the second point of intersection of $(c)$ and $\left(c_{1}\right)$. First we will prove that the lines $D_{1} H_{2}, D_{2} H_{3}$ and $U D$ pass through the point $L$.

Suppose that the line $A H$ intersects the side $B C$ at the point $Z$. Since $H_{1} D\|B C\| D_{1} D_{2}$ and $B, C$ are the midpoints of the segments $D_{1} D, D_{2} D$ respectively, we get that $Z$ is the midpoint of the segment $H H_{1}$, so the point $H$ lies on $D_{1} D_{2}$. Therefore, $A H \perp D_{1} D_{2}$ and $A U$ is a diameter of $\left(c_{1}\right)$. Thus, $A L \perp U L$ and $A L \perp D L$. We have that the points $U, D, L$ are collinear. (1)

Now, $\angle A L D_{1}=\angle A D_{2} D_{1}, \angle A L H_{2}=\angle A C H_{2}$. Since $A H C D_{2}$ is cyclic we get
$\angle A C H_{2}=\angle A D_{2} D_{1}$. Therefore, $\angle A L H_{2}=\angle A L D_{1}$. So the points $D_{1}, H_{2}, L$ are collinear. (2)

Similarly,

$$\angle D_{1} L D_{2}=\angle D_{1} A D_{2}=180^{\circ}-2\left(\angle A D_{1} H\right) .$$

Since $A D_{1} B H$ is cyclic we have $\angle A D_{1} H=\angle A B H=\angle A B H_{3}$. Therefore, we get

$$\angle D_{1} L D_{2}=180^{\circ}-2\left(\angle A B H_{3}\right)=180^{\circ}-2\left(\angle A D H_{3}\right)=180^{\circ}-\angle H_{2} D H_{3} .$$

Thus,

$$\angle D_{1} L D_{2}+\angle H_{2} D H_{3}=180^{\circ} \quad \text { or } \angle D_{1} L D_{2}+\angle H_{3} L H_{2}=180^{\circ} .$$

So the points $H_{3}, L, D_{2}$ are collinear. (3)
From (1), (2), (3) we have that the lines $D_{1} H_{2}, D_{2} H_{3}$ and $U D$ are concurrent at the point $L$.

Also we have

$$\angle H_{3} D A=\angle D_{2} D A-\angle C D H_{3}=\angle A D_{2} D-\angle C B H_{3}$$

and because $B \mathrm{HD}_{2} \mathrm{C}$ is a parallelogram, we get $\angle C B H_{3}=\angle H D_{2} C$. So

$$\angle H_{3} D A=\angle A D_{2} D-\angle H D_{2} C=\angle A D_{2} D_{1}=\angle A D_{1} D_{2}=\angle A U D_{2} .$$

Therefore, the circumcircle of the triangle $U D L_{1}$ passes through the point $A$. Also, $\angle A D_{1} K_{1}=\angle D_{2} D_{1} A=\angle D_{2} U A$. But $A U L_{1} D$ is cyclic and we have $\angle D_{2} U A=\angle H_{3} D A=$ $\angle H_{3} B A=\angle H_{3} H_{2} A$. Therefore, $\angle A D_{1} K_{1}=\angle H_{3} H_{2} A$. Thus, the circumcircle of the triangle $D_{1} K_{1} H_{2}$ passes through the point $A$.

Because the points $H_{3}, L, D_{2}$ are collinear by the Desargues theorem, the lines $U D_{1}$, $L_{1} K_{1}, D H_{2}$ are concurrent, let say in the point $M$.

From the similarity of the triangles $U D L_{1}$ and $D_{1} K_{1} H_{2}$ we conclude that $M$ is the center of unique spiral similarity and because the circumcircles of the triangles $D_{1} K_{1} H_{2}$ and $U D L_{1}$ intersect at the point $A$, then the second point of intersection is $M$. Therefore, $M$ lies on the line $K_{1} L_{1}$.

Comment. We can prove the last part in a different way.
Let $M$ be the point of intersection of the circumcircles of the triangles $D_{1} K_{1} H_{2}$ and $U D L_{1}$. Now, we have

$$\angle K_{1} M A=\angle H_{3} H_{2} A=\angle H_{3} B A=\angle A D H_{3}=\angle L_{1} U A=\angle L_{1} M A .$$

Therefore, the points $L_{1}, K_{1}, M$ are collinear.