$ABCD$ is a cyclic quadrilateral inscribed of circle $O$. The tangent line of circle $O$ at $B$ intercepts $CD$ at $K$. The tangent line of circle $O$ at $C$ intercepts $AB$ at $M$.
if $AB = BM$ and $CD=CK$, prove that $ABCD$ is a trapezoid.
I was going down the route of:
$\frac{BM}{\sin \angle BCM} = \frac{BC}{\sin \angle BMC}$, so $BM = \frac{BC \sin \angle BCM}{\sin \angle BMC}$
$\frac{AB}{\sin \angle ACB} = \frac{BC}{\sin \angle BAC}$, so $AB = \frac{BC \sin \angle ACB }{\sin \angle BAC}$
and $\angle BCM = \angle BAC$
$\frac{BM}{AB} = \frac{\sin \angle BCM \sin \angle BAC}{\sin \angle BMC \sin \angle ACB} = 1$
but the relationship is awkward here.
It's enough to prove that $BD=AC$.
Observe that, $\triangle BCK\sim \triangle DBK$ and $\triangle CBM\sim \triangle ACM$.
Hence, $\frac{BK}{DK}=\frac{CK}{BK}$
$\Rightarrow BK=CK\sqrt{2}$
Thereafter, $\frac{BD}{BC}=\frac{BK}{CK}=\sqrt{2}$
Similarly yield $\frac{AC}{BC}=\sqrt{2}$ and thus $BD=BC\sqrt{2}=AC$.