Let $ABCD$ be a cyclic quadrilateral with circumcircle $\Gamma$. Let $M$ be the midpoint of the arc $ABC$. The circle with center $M$ and radius $MA$ meets $AD, AB$ at $X, Y$. The point $Z \in XY$ with $Z \neq Y$ satisfies $BY=BZ$. Show that $\angle BZD=\angle BCD$.
We have that $DM$ is the angle bisector of $\angle ADC$ and that $BM$ is the exterior angle bisector of $\angle ABC$, so from $MX=MC=MY$ we get that $X$ and $C$ are symmetric wrt $MD$ and that $Y$ and $C$ are symmetric wrt $BM$. Let $Z'$ be the reflection of $C$ wrt $DB$. Then $X,Y,Z'$ lie on the Steiner Line of $C$ wrt $BMD$ and $\angle BZ'D=\angle BCD$. Hence $Z=Z'$, so $BY=BC=BZ$ (and also $DX=DC=DZ$). Done.
[asy][asy]
/* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */
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[/asy][/asy]
For conciseness we define \(\beta = \angle ABC\).
We start with a "lemma".
\(BY = BC\)
This is fairly well known but I'll include it for completeness, we have
\[\angle BYC = 180 - \angle CYA = \dfrac{1}{2}\angle CMA = 90 - \dfrac{\beta}{2}\]However we have \(\angle YBC = \beta\) whence the conclusion immediately follows
We wish to prove that \(\triangle BZD \cong \triangle BCD\), this clearly finishes the problem and due to the
previous lemma it suffices to show that \(\angle CBD = \angle DBZ\). We shall chase \(\angle BYZ\)
\begin{align*}
\angle BYZ &= \angle BYX \\
&= \angle BYC + \angle CYX \\
&= 90 - \dfrac{\beta}{2} + \angle CAD && \tag{\(CYAX\) is cyclic} \\
&= 90 - \dfrac{\beta}{2} + \angle CBD \\
\end{align*}We thus obtain that
\[\angle ZBY = 180 - 2\angle BYZ = \beta - 2\angle CBD\]However it is clear that
\[\angle ZBC = \beta - \angle ZBY = 2\angle CBD\]We now obtain
\[\angle DBZ = \angle ZBC - \angle CBD = \angle CBD\]Thus we are done, we obtain that \(\triangle BZD \cong \triangle BCD\) and thus \(\angle BZD = \angle BCD\).