Proof of Incircle

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.

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 $PQR$.