Solution:

Let $\Gamma$ be the circle centered at $A$ and passing through $B, C, D$. Let $P$ also be any point on the arc $BD$ not containing $C$. Since the inscribed angle $\widehat{BPD}$ subtends the same arc as the central angle $\widehat{BAD}$, we have $\widehat{BAD} = 2 \cdot \widehat{BPD}$; on the other hand, $\widehat{BAD} = 2 \cdot \widehat{BAM}$ by construction, so $\widehat{BPD} = \widehat{BAM}$. Moreover, the angles $\widehat{BPD}$ and $\widehat{BCD}$ subtend the same arc, but from opposite sides: they are therefore supplementary. It follows that the angles $\widehat{BAM}$ and $\widehat{BCM}$ are supplementary, that is, that the quadrilateral $AMBC$ can be inscribed in a circle, which we shall call $\gamma$.

The quadrilateral $AMCB$ is inscribable in a circle if and only if $\widehat{BAM} + \widehat{BCM} = 180^\circ$, that is, if and only if $2 \cdot \widehat{BAM} + 2 \cdot \widehat{BCM} = 360^\circ$. Now, $2 \cdot \widehat{BAM} = \widehat{BAD}$ by construction; on the other hand, $\widehat{BCD} = \widehat{BCA} + \widehat{ACD} = \widehat{CBA} + \widehat{CDA}$. The claim is therefore equivalent to $\widehat{BAD} + \widehat{BCD} + \widehat{CBA} + \widehat{CDA} = 360^\circ$, which is true since the sum contains precisely the interior angles of the quadrilateral $ABCD$.

b.
$\widehat{BMA} = \widehat{BCA}$ since they both subtend the arc $AB$ of $\gamma$. Moreover $\widehat{BCA} = \widehat{CBA}$ since they are the base angles of triangle $ABC$, which is isosceles by hypothesis. The triangles $ABM$ and $ANB$ therefore have $\widehat{BMA} = \widehat{NBA}$ by what has just been proven, and the angle at $\widehat{A}$ in common, so they are similar by the second similarity criterion.