We claim that $L$ lies on $(AED)$ and $(CDF)$. Redefine $L$ to be the second intersection of $(AED)$ and $(CDF)$. [asy][asy] size(8cm); pair A = (-1, 0), B = (1, 0), C = dir(85), D = foot(C, A, B), E = rotate(60, A)*D, F = rotate(60, C)*D, L = extension(A, C, E, F); draw(A--B--C--cycle, red); draw(A--E--D^^C--D--F--cycle, magenta); draw(circumcircle(A, E, D)^^circumcircle(C, D, F), orange); draw(E--F, mediumred+dashed); string[] names = {"$A$", "$B$", "$C$", "$D$", "$E$", "$F$", "$L$"}; pair[] points = {A, B, C, D, E, F, L}; pair[] ll = {A, B, C, dir(-80), E, F, L}; for (int i=0; i<names.length; ++i) dot(names[i], points[i], dir(ll[i])); [/asy][/asy] Since \[ \angle ELD = \angle EAD = 60^\circ= \angle FCD = \angle FLD, \]$L$ lies on line $EF$. Similarly, since \[ \angle DLA = 60^\circ= \angle DFC = \angle DLC, \]$L$ lies on line $AC$. Thus, $L$ is as described, and we're done by van Schooten's theorem.

A $60^\circ$ rotation about $D$ sends $A$ to $E$ and $C$ to $F$, thus $\triangle EDF\cong\triangle ADC$ and $\widehat{ALE}=60^\circ$, $L$ is the common point of the circles $\odot (AED)$ and $\odot (CDF)$, next as above. Remark: there is no need of $AC\bot BC$. Best regards, sunken rock

Pretty easy Just notice that the spiral sim centred at $D$ takes $AE \longrightarrow BF$ which means that $D=\odot(LAE) \cap \odot(LBF)$. Van schooten’s theorem finishes $\blacksquare$