Points $A, B$, and $C$ are on a line in this order. Points $D$ and $E$ lie on the same side of this line, in such a way that triangles $ABD$ and $BCE$ are equilateral. The segments $AE$ and $CD$ intersect in point $S$. Prove that $\angle ASD = 60^o$. [asy][asy] unitsize(1.5 cm); pair A, B, C, D, E, S; A = (0,0); B = (1,0); C = (2.5,0); D = dir(60); E = B + 1.5*dir(60); S = extension(C,D,A,E); fill(A--B--D--cycle, gray(0.8)); fill(B--C--E--cycle, gray(0.8)); draw(interp(A,C,-0.1)--interp(A,C,1.1)); draw(A--D--B--E--C); draw(A--E); draw(C--D); draw(anglemark(D,S,A,5)); dot("$A$", A, dir(270)); dot("$B$", B, dir(270)); dot("$C$", C, dir(270)); dot("$D$", D, N); dot("$E$", E, N); dot("$S$", S, N); [/asy][/asy]