An isosceles trapezoid $ABCD$ with bases $BC$ and $AD$ is given. Circles with centers $O_1$ and $O_2$ are inscribed in triangles $ABC$ and $ABD$. Prove that line $O_1O_2$ is perpendicular on $BC$.
Let the midpoints of $BC$ and $AD$ be $M$ and $N$ respectively. Let the incircle of triangle $ABC$ touch $BC$ at $P_1$, and incircle of triangle $ABD$ touch $AD$ at $P_2$.
In order to show that $O_1O_2\perp BC$, it suffices to show that $P_1M=P_2N$, as this implies that $P_1MP_2N$ is a parallelogram, so $O_1O_2||MN$, so $O_1O_2\perp BC$ follows from the fact that $MN\perp BC$.
This is simply length chasing:
\begin{align*}
P_1M&=BM-BP_1\\
&=\frac12 BC-\frac12(AB+BC-AC)\\
&=-\frac12 AB+\frac12 AC
\end{align*}\begin{align*}
P_2N&=AN-AP_2\\
&=\frac12 AD-\frac12(AB+AD-BD)\\
&=-\frac12 AB+\frac12 BD\\
&=-\frac12 AB+\frac12 AC\end{align*}Therefore, $P_1M=P_2N$, which is what we wanted to prove.