$SE \perp BD$. $AC$ is tangent to $(BCD)$. Hence, $\angle SCA=90$ Hence $AESC$ is a cycic quadrilateral.

How do you know $SE\perp BD$?

And also how do you know that S is located in BC?

@Tedi_K._10 I explained in more depth why $SE\perp BD$, but it is not true that $S$ is on $BC$...

I think this is an easier proof : Let angle CBD be x and let angle DCB = y, then we have that BDC = 180 - x - y. Since S is the circuncenter BCD, angle BSC (the exterior one) = 2 angle BDC = 360 - 2x - 2y, then angle BSC(the interior one) = 2x + 2y. As BSC is an isosceles triangle, angle SBC = angle SCB = 90 - x - y. And we have that ACS = ACD + DCB + BCS = x + y + (90 - x - y) = 90. As we want to show that AESC is cyclic, it's enough to prove that SEA = 90 but this is true since SE is the perpendicular bissector of BD.

We know that $SD = SB = SC$, as $SE$ is median of the isosceles triangle $SDB$, then $SE$ is height and $\angle SEA = 90º$, too $\angle ACD = \angle DBC$, thus $AC$ is tangent at $C$ to the circumcircle of $BCD \implies \angle SCA = 90º$. Therefore $\angle SEA + \angle SCA = 180º.$ So $A, E, S, C$ lie on a circle.

The angle condition implies $\odot (BCD)$ tangent to $AC$ $\implies$ $\angle ACS$ $=$ $90^{\circ}$ $=$ $\angle AES$ $\implies$ $ACSE$ concyclic