Problem: Let $ABC$ be a triangle and let $\omega$ be its incircle.
Denote by $D_1$ and $E_1$ the points where $\omega$
is tangent to sides $BC$ and $AC$, respectively.
Denote by $D_2$ and $E_2$ the points on sides $BC$ and $AC$,
respectively, such that $CD_2=BD_1$ and $CE_2=AE_1$,
and denote by $P$ the point of intersection of segments $AD_2$ and $BE_2$.
Circle $\omega$ intersects segment $AD_2$ at two points,
the closer of which to the vertex $A$ is denoted by $Q$.
Prove that $AQ=D_2P$.

Solution: We have that $P$ is the Nagel point
\[ P = \left( s-a : s-b : s-c \right). \]
Therefore,
\[ \frac{PD_2}{AD_2} = \frac{s-a}{(s-a)+(s-b)+(s-c)} = \frac{s-a}{s}. \]
Meanwhile, $Q$ is the antipode of $D_1$.
The classical homothety at $A$ mapping $Q$ to $D_1$
(by mapping the incircle to the $A$-excircle)
has ratio $\frac{s-a}{s}$ as well
(by considering the length of the tangents from $A$),
so we are done.