Angle Bisectors A circle is drawn that intersects all three sides of $\triangle PQR$ as shown below. Prove that if $AB = CD = EF$, then the center of the circle is the incenter of $\triangle PQR$.

 A: Let $Q'$ be the midpoint of $FE$, $R'$ be the midpoint of $AB$ and $P'$ be the midpoint of $CD$. The perpendicular bisectors of $FE, AB, CD$ meet at the centre of our circle $\Gamma$, and assuming $AB=CD=EF=2l$ we have
$$\text{pow}_{\Gamma}(P) = PA\cdot PB = PF\cdot PE = PR'^2-l^2 = PQ'^2-l^2$$
hence $PQ'=PR'$ and in a similar way $QP'=QR'$ and $RP'=RQ'$. 
So we have that $P',Q',R'$ are the contact points of the inscribed circle and the centre of $\Gamma$ is the incenter of $PQR$.
Since I believe the question will be marked as a duplicate soon, I posted this solution at the original thread, too.
A: Take a circle and an intersecting line. Make the radius of the circle shrink, then by the symmetry of the circle, the chord will shrink symmetrically to the bisecting line (its axis). Always by the circle symmetry, any other chord of same (original) length will shrink in the same way and by the same amount.
So they will altogether become tangent to circle at their center point, i.e. at the respective bisecting line, i.e. at the normal to their central point.
