Excircle and incircle proof 
Prove that if the incircle of triangle $ABC$ touches side $BC$ at $D$ and the $A$-excircle touches side $BC$ at $D'$, then the midpoint of $BC$ is the midpoint of $DD'$.

This is an interesting property that I discovered when doing a few problems but the solutions didn't prove it. After drawing several triangles and their in- and excircles, it seems to be true that the midpoint of the intouch and and extouch points is in fact the midpoint of the side of the triangle. 
As another question, if anyone can prove that the antipode of the intouch $D$ which I have labeled as $U$ in the triangle below is collinear with the exsimilicenter $A$ and $D'$ that would help. Finally, is it also true that $A,U,D',$ and $I_a$ are collinear or just $A,U,$ and $D'$? 

Note: $I_a$ is the center of the $A$-exircle.

I will answer the first question.
This is easy to prove using just one basic idea: when a circle is tangent to two sides of an angle, the distance from the vertex to each of the points of tangency is the same.
Applying that idea to the incircle, you'll find after some calculations that $BD = \frac{1}{2}(a + c - b)$. Applying it to the excircle opposite vertex $A$, you'll find $CD' = \frac{1}{2}(a + c - b) = BD$.
For the second question, the dilation $h$ with centre $A$ taking $I$ to $I_a$ transforms the incircle into the excircle. It also transforms line $ID$ into a parallel line passing through $I_a$. Since $ID$ and $I_aD'$ are both perpendicular to $BC$, the image of line $ID$ must be line $I_a D'$. Points $U$ and $D$ are the intersections of the incircle with $ID$. They must be carried by $h$ to the intersections of the excircle with $I_a D'$. Since $h$ has positive ratio, $U$ must be carried to $D'$. This proves that $A, U, D'$ are aligned. 
Assume $I_a$ is on line $AD'$. Then since $I_a D'$ and $UD$ are both perpendicular to $BC$ and pass through $U$, we must have $D = D'$. This shows that $AI$ is is perpendicular to $BC$, hence that $ABC$ is isosceles.
