Proof of Incircle 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 $\triangle PQR$. 

Designate the center of the circle $G$.
Thinking about it a bit, we realize if we could prove the incircle of the triangle and the circle given are concentric, we would know that the center of the circle is the incenter, for they have the same center.  
If we shrink the circle given down to the incircle, the incircle should intersect the triangle at the midpoints of $AB$, $CD$, and $EF$.  
We therefore denote the midpoint of $AB M$, the midpoint of $CD L$, andd the midpoint of $EF N$.
We know the incircle is concentric to the given circle if and only if $NG$ and $LG$ are congruent, because a circle is a circle if and only if the radii are congruent.  Thus, we must prove $NG\cong LG$.
When proving lengths equal, congruent triangles are always a good idea.  We see that $NGE\cong LGD$ seems likely, and it would also prove our answer if we could prove it. 
We know that $GD\cong GE$ because they are both radii of the larger circle, and we also know $DL\cong EN$ because they are both half of two lines that are congruent.
We know $\angle GLD$ and $\angle GNE$ are both right angles, because that is where the smaller circle is tangent to the sides of the triangle.  Thus, $NGE\cong LGD$ by $HL$ congruency.
Because congruent parts of congruent triangles are congruent, we know $NG\cong LG$, which proves $G$ is the center of the incircle.
Is this proof sound?
 A: Your solution is great but a bit heavy on notation. It reminds me of reading Euclid, which is not bad but still a bit heavy. Here is my version:

Since the three chords have the same length, the orthogonal heights of the three circular arcs must be equal. Call these heights $h$ and it follows immediately, that if the radius of the intersecting circle is $r$ the center $G$ lies at a distance of $r-h$ from each of the three sides of the triangle.
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 $PQR$.
A: I would develop your point just using circle symmetry.
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.
