Spoiler:
Let's call the vertices of the triangle A, B, and C. And we'll call four of the extended points D, E, F, and G. I've also labeled the measures of the angles of the triangle: M_a, M_b, and M_c.
Notice that CD and EC have the same total length. This means that triangle CED is an isosceles triangle and angles CDE and DEC are equal to each other (as marked in the diagram). But we also know that the sum of the angles of triangle CED is 180 degrees. So:
180 = <DCE + <DEC + <CDE
180 = M_c + 2*<DEC
<DEC = <CDE = 90 - 0.5*M_c
Similarly, we can find that:
<BFG = <BGF = 90 - 0.5*M_b
Next, notice that triangle ADG is also isosceles, so <ADG = <AGD. Also we can see that <DAG and <BAC are equivalent angles. So <DAG = M_a and that means:
<ADG = <AGD = 90 - 0.5*M_a
Finally, calculate the sum of <FGD + <BED:
[<FGD] + [<BED]
= [(<BGF) + (<AGD)] + [<CED]
= [(90 - 0.5*M_b) + (90 - 0.5*M_a)] + [90 - 0.5*M_c]
= 270 - 0.5*M_a - 0.5*M_b - 0.5*Mc
= 270 - 0.5*(M_a + M_b + M_c)
= 270 - 0.5*180 = 270 - 90 = 180
So <FGD + <BED = 180 and it can be similarly shown that <EDG + <CFG = 180 as well. This means that quadrilateral DEFG is cyclic, so it's vertices must all fall on the same circle.
And this same method can be used to show that the other two extended points fall on the same circle as well.