[asy][asy] pair A = dir(60); pair B = dir(210); pair C = dir(330); pair D = 0.5B + 0.5C; pair E = foot(D, A, B); pair O = circumcenter(A, B, C); pair Z = extension(A, O, D, E); draw(circumcircle(C, A, B), mediumblue+dashed); draw(C--A--B--cycle, heavyblue); draw(D--E, deepgreen); draw(A--O--Z, deepgreen); draw(circumcircle(C, A, Z), lightblue+dotted); dot("$A$", A, dir(A)); dot("$B$", B, dir(B)); dot("$C$", C, dir(C)); dot("$O$", O, (1.5,0)); dot("$D$", D, dir(D)); dot("$E$", E, dir(E)); dot("$Z$", Z, dir(Z)); [/asy][/asy]

We have: $\widehat{BDZ}$ = $90^o$ $-$ $\widehat{ABC}$ = $\widehat{CAO}$ So: $A$, $C$, $D$, $Z$ lie on a circle

There was no need of the condition $D$ is the midpoint of $BC$. Inverted just for fun... Greek JBMO TST 2019 P1 wrote: Consider an acute triangle $ABC$ with $AB>AC$ inscribed in a circle of center $O$. From the midpoint $D$ of side $BC$ we draw line $(\ell)$ perpendicular to side $AB$ that intersects it at point $E$. If line $AO$ intersects line $(\ell)$ at point $Z$, prove that points $A,Z,D,C$ are concyclic. After Inversion around $A$ with any arbitary radius the problem is read as follows. Inverted Problem wrote: In a circle there are four arbitary points $A,B',C',D'$. Let the line perpendicular to $AD'$ at $D'$ intersect $AB'$ at $E'$. Let the line perpendicular to $B'C'$ through $A$ intersect $\odot(AD'E')$ at $Z'$. Then $\overline{D'-Z'-C'}$. $\angle AD'Z'=\angle AE'Z'=\angle AB'C'=\angle AD'C'$. Hence, Inverting back we get that $A,D,Z,C$ are concyclic.

$\angle AZD=180 -\angle AZE=90+\angle BAO=180-\angle C$

Connect OA and OC $\angle OAC=\angle OCA=90-B$ $\angle AZD+\angle ACD=90+A-90+B+C=180$ and we are done

∠ZAC = 90 - ∠B = ZDB ---> ZACD is cyclic.