10.7 As shown in Figure 4, let the incircle of $\triangle ABC$ touch its sides at points $A_{1}, B_{1}, C_{1}$. Draw a line $a_{1}$ through $A_{1}$ parallel to the angle bisector of $\angle A$. Since $\triangle AB_{1}C_{1}$ is an isosceles triangle, the angle bisector of $\angle A$ is perpendicular to $B_{1}C_{1}$, thus $a_{1} \perp B_{1}C_{1}$, meaning $a_{1}$ passes through the altitude of $\triangle A_{1}B_{1}C_{1}$ on side $B_{1}C_{1}$.

As shown in Figure 5, let the midpoints of the sides of $\triangle ABC$ be $A_{0}, B_{0}, C_{0}$. Since $\triangle A_{0}B_{0}C_{0}$ is similar to $\triangle ABC$ with a similarity ratio of $\frac{1}{2}$, the angle bisector of $\angle A$ is parallel to the angle bisector of $\angle A_{0}$. Let the incenter of $\triangle A_{0}B_{0}C_{0}$ be $S$.

It is well known that point $A^{\prime}$ is equidistant from point $A_{1}$ and the midpoint of side $BC$ (the same property holds for point $B^{\prime}$ with point $B_{1}$, and point $C^{\prime}$ with point $C_{1}$).

Perform a symmetry transformation about point $S$. In this transformation, line $a_{1}$ becomes line $a$. Thus, under this transformation, the altitudes of $\triangle A_{1}B_{1}C_{1}$ are transformed to lines $a, b, c$, respectively, so these three lines intersect at a single point, which is the reflection of the orthocenter of $\triangle A_{1}B_{1}C_{1}$ about $S$.