Elegant proof
Let's start with a generic cyclic quadrilateral, and define
\begin{align*}
\angle ACB = \angle ADB &= \alpha \\
\angle CAD = \angle CBD &= \beta \\
\angle DBA = \angle DCA &= \gamma \\
\angle BAC = \angle BDC &= 180°-\alpha-\beta-\gamma
\end{align*}
This answer is based on a key observation: $K$ is the center of the circle through $B,C,M$.
This will certainly ensure that $BCK$ is isosceles with base $BC$. We'll check the angle sum condition along the way. If $K$ is the center of the circumcircle, then not only $BCK$ bit also $BMK$ and $CMK$ are isosceles, and we have the following conditions (introducing a new angle $\delta$ which we'll atempt to eliminate):
$$
\angle KBC=\angle BCK=\delta \\
\angle KMC=\angle MCK=\alpha+\delta \\
\angle BMK=\angle KBM=\beta+\delta \\
\angle CKM=180°-2(\alpha+\delta) \\
\angle MKB=180°-2(\beta+\delta) \\
\angle CKB=180°-2\delta=
\bigl(180°-2(\alpha+\delta)\bigr)+\bigl(180°-2(\beta+\delta)\bigr)
$$
From his last equation you can conclude
$$\delta=90°-\alpha-\beta$$
So now we can verify that
$$
\angle KBC+\angle AMB = \delta+(180°-\angle BAC-\angle DBA)\\
= 90°-\alpha-\beta+(180°-(180°-\alpha-\beta-\gamma)-\gamma)=90°
$$
so $K$ is indeed the right point. Now consider the quadrilateral $ABKH$, where $H$ is the intersection of $KM$ with $AD$. It has interior angles
\begin{align*}
\angle BAH &= \angle BAC + \angle CAD
= (180°-\alpha-\beta-\gamma)+\beta = 180°-\alpha-\gamma \\
\angle KBA &= \angle KBC + \angle CBD + \angle DBA
= (90°-\alpha-\beta) + \beta + \gamma = 90°-\alpha+\gamma \\
\angle HKB &= 180°-2(\beta+\delta) = 2\alpha \\
\angle AHK &= 360°-\angle BAH-\angle KBA-\angle HKB
= 360°-180°+\alpha+\gamma-90°+\alpha-\gamma-2\alpha = 90°
\end{align*}
$$\tag*{$\Box$}$$
Brute force proof
The above proof relies on spotting that fact about the circumcircle. Before I had that, I had a less elegant proof by brute force computation using sage. In fact, I found out that $BMK$ and $CMK$ are isosceles by comparing angles obtained from the construction below for random parameters. That computation is based on the following construction:
Start with four points on the unit circle, using homogeneous coordinates and a rational parametrization.
P.<a, b, c, d> = ZZ[]
def cpt(u): # a point on the unit circle
return vector([1-u^2, 2*u, 1+u^2])
A = cpt(a)
B = cpt(b)
C = cpt(c)
D = cpt(d)
$$
A = \begin{pmatrix}1-a^2\\2a\\1+a^2\end{pmatrix} \qquad
B = \begin{pmatrix}1-b^2\\2b\\1+b^2\end{pmatrix} \qquad
C = \begin{pmatrix}1-c^2\\2c\\1+c^2\end{pmatrix} \qquad
D = \begin{pmatrix}1-d^2\\2d\\1+d^2\end{pmatrix}
$$
You can join points and intersect lines using the cross product.
def join(a, b):
v = a.cross_product(b)
g = gcd(v)
return v/g
meet = join
M = meet(join(A, C), join(B, D))
$$ M=\begin{pmatrix}
a b c - a b d + a c d - b c d - a + b - c + d \\
-2 a c + 2 b d \\
- a b c + a b d - a c d + b c d - a + b - c + d
\end{pmatrix} $$
Next, construct the line through $M$ orthogonal to $BD$ and intersect this with $CD$ to obtain $E$. This will ensure $\angle AMB+\angle EMC=90°$.
perp = diagonal_matrix([1, 1, 0])
E = meet(join(perp*join(B, D), M), join(C, D))
$$ E=\begin{pmatrix}
a b^{2} c^{2} d^{2} - a b^{2} c d^{3} + a b c^{2} d^{3} - b^{2} c^{2}
d^{3} + a b^{2} c^{2} + a b^{2} c d - a b c^{2} d \\{}- b^{2} c^{2} d - 2
a b^{2} d^{2} + 2 a b c d^{2} - b^{2} c d^{2} + b c^{2} d^{2} - a b
d^{3} + 2 b^{2} d^{3} \\{}+ a c d^{3} - 2 b c d^{3} - 2 a b c + b^{2} c + 2
a c^{2} - b c^{2} + a b d - a c d \\{}+ 2 b c d - 2 c^{2} d - a d^{2} -
b d^{2} + c d^{2} + d^{3} - a + b - c + d \\[2ex]
{}-2 a b c^{2} d^{2} + 2 b^{2} c d^{3} - 2 a b^{2} c + 2 a b^{2} d - 4 a b
c d + 4 b^{2} c d + 2 a c^{2} d \\{}- 2 b c^{2} d + 2 a b d^{2} - 2 b^{2}
d^{2} - 4 a c d^{2} + 4 b c d^{2} - 2 c^{2} d^{2} + 2 c d^{3} - 2 a c +
2 b d \\[2ex]
{}- a b^{2} c^{2} d^{2} + a b^{2} c d^{3} - a b c^{2} d^{3} + b^{2} c^{2}
d^{3} - a b^{2} c^{2} + a b^{2} c d - a b c^{2} d \\{}+ b^{2} c^{2} d - 2
a b c d^{2} + b^{2} c d^{2} - b c^{2} d^{2} + a b d^{3} - a c d^{3} +
2 b c d^{3} - 2 a b c \\{}+ b^{2} c - b c^{2} + a b d - a c d + 2 b c d -
a d^{2} + b d^{2} - c d^{2} + d^{3} - a + b - c + d
\end{pmatrix} $$
Compute $F$ as the other intersection of $BC$ with the circle through $CEF$. Due to the inscribed angle theorem, this will give you $\angle EFC$ = $\angle EMC$.
def cvec(p):
x, y, z = p
return vector([x*x + y*y, x*z, y*z, z*z])
P2.<mu> = P[]
p = list(matrix(map(cvec, [C, E, M, B+mu*C])).det())
F = p[1]*B - p[0]*C
F = F/gcd(F)
$$ F=\begin{pmatrix}
{}- a b^{4} c^{2} + a b^{4} c d - a b^{3} c^{2} d + b^{4} c^{2} d - 2 a
b^{3} c + b^{4} c + a b^{2} c^{2} - b^{3} c^{2} \\{}+ 3 a b^{3} d - 2 b^{4}
d - 2 a b^{2} c d + 2 b^{3} c d - a b c^{2} d + b^{2} c^{2} d + a b^{2}
- b^{3} + 2 a b c \\{}- 2 b^{2} c - 2 a c^{2} + 3 b c^{2} - a b d + b^{2}
d + a c d - 2 b c d + a - b + c - d \\[2ex]
2 a b^{3} c^{2} - 2 b^{4} c d + 6 a b^{2} c - 4 b^{3} c - 2 a b c^{2} \\{}+
4 b^{2} c^{2} - 4 a b^{2} d + 2 b^{3} d + 4 a b c d - 6 b^{2} c d + 2 a
c - 2 b d \\[2ex]
a b^{4} c^{2} - a b^{4} c d + a b^{3} c^{2} d - b^{4} c^{2} d + 2 a
b^{3} c - b^{4} c + a b^{2} c^{2} \\{}+ b^{3} c^{2} - a b^{3} d - 2 b^{3}
c d + a b c^{2} d - b^{2} c^{2} d + a b^{2} - b^{3} \\{}+ 2 a b c + b
c^{2} - a b d - b^{2} d + a c d - 2 b c d + a - b + c - d
\end{pmatrix} $$
Let $G$ be the midpoint of $BC$.
def midpoint(a, b):
v = b[-1]*a + a[-1]*b
g = gcd(v)
return v/g
G = midpoint(B, C)
$$ G=\begin{pmatrix}
- b^{2} c^{2} + 1 \\
b^{2} c + b c^{2} + b + c \\
b^{2} c^{2} + b^{2} + c^{2} + 1
\end{pmatrix} $$
Then you can construct the line parallel to $EF$ through $B$ and intersect this with the radius $OG$ to obtain $K$. The parallel line ensures $\angle KBC=\angle EFC$.
linf = vector([0, 0, 1])
O = vector([0, 0, 1])
K = meet(join(meet(join(E, F), linf), B), join(O, G))
$$ K=\begin{pmatrix}
- a b c + b c d + a - d \\
a b + a c - b d - c d \\
a b c - a b d + a c d - b c d + a - b + c - d
\end{pmatrix} $$
Now verify that the lines $KM$ and $AD$ are indeed orthogonal
join(K, M)[:2].dot_product(join(A, D)[:2]).is_zero()
This will print the desired answer, True.
