Let O be the center of the regular hexagon with side a=BC. We know that triangles OAB, OBC, OCD, ODE, OEF, OFA are equilaterals with side a=BC. BP bisects angle <OBC= 60^o because OBCD is rhombus , so <OBD=30^o similarly <ODQ=30^o Draw circle with center B and radius a=BC=BO, then <OPC=(QP)/2= <BQP = 30^o Draw circle with center D and radius a= DE= DO=DC, then <OCQ=(OQ)/2 = <OQD= 30 ^o so <OCQ = 30^o= <OPC which means that O, P, Q are collinear Notation: (XY) stands for small arc XY

[asy][asy] pair A=(0,2sqrt(3)); pair B=(2,2sqrt(3)); pair C=(3,sqrt(3)); pair D=(2,0); pair E=(0,0); pair F=(-1,sqrt(3)); draw(A--B--C--D--E--F--cycle); label("$A$",A,NW); label("$B$",B,NE); label("$C$",C+(0.23,0),E); label("$D$",D,SE); label("$E$",E,SW); label("$F$",F,W); draw(B--D); draw(D--F); pair P=(2,2sqrt(3)-2); pair Q=(2-sqrt(3),1); label("$P$",P,NW); label("$Q$",Q,NE); dot(P); dot(Q); draw(C--Q); add(pathticks(D--E,2,.5,0,10)); add(pathticks(D--Q,2,.5,0,10)); add(pathticks(B--P,2,.5,0,10)); [/asy][/asy] WLOG $s=2$, let $E=(0,0).$ We have $D=(2,0)$, $F=(-1,\sqrt3)$, $B=(2,2\sqrt3),$ and $C=(3,\sqrt3)$. Note that the slope of $DF$ is $-\frac{1}{\sqrt3}.$ $P$ is $(2,2\sqrt3-2)$ since you just go down $2$ from $B$. $Q$ is $(2-\sqrt3,1)$ since from $D$, you go to the left $\sqrt3$ and up $1$, which is a distance of $2$ and matches the slope of $DF.$ We can now see that $C$, $P$, and $Q$ are collinear, since they all pass through the line $y=(-\sqrt3+2)x+(4\sqrt3-6).$