Let $O$ be the centre of the circle and $r$ be its radius. Draw $OE \perp AB$, $OF \perp BD$ and $OG \perp DC$. Then $|AE| = r \cos \angle BAO$. Noting that $\angle BAO = \angle CAO$ and $E$ is the midpoint of $AB$, it follows that
$$|AB| + |AC| = 4r \cos \angle BAO.$$

Using $|DF| = r \cos \angle BDO$ and $|DG| = r \cos \angle CDO$, we get
$$\begin{aligned} |DB| + |DC| &= 2r \cos \angle BDO + 2r \cos \angle CDO \\ &= 2r(\cos \angle BDO + \cos \angle CDO) \\ &= 4r \cos \left( \frac{\angle BDO + \angle CDO}{2} \right) \cos \left( \frac{\angle BDO - \angle CDO}{2} \right) \end{aligned}$$
Because $4r \cos(\frac{\angle BDC}{2}) = 4r \cos(\frac{\angle BAC}{2}) = 4r \cos \angle BAO = |AB| + |AC|$, we obtain
$$|DB| + |DC| = (|AB| + |AC|) \cos \left( \frac{\angle BDO - \angle CDO}{2} \right)$$
Since $\angle BDC < 180^\circ$, $\angle BDO < 90^\circ$ and $\angle CDO < 90^\circ$ then
$$-90^\circ < \frac{\angle BDO - \angle CDO}{2} < 90^\circ \Rightarrow 0 < \cos \left( \frac{\angle BDO - \angle CDO}{2} \right) < 1$$
with equality on the right iff $\angle BDO = \angle CDO$. Because $\angle BDO = \angle CDO$ is equivalent to $A = D$, we obtain $|DB| + |DC| \le |AB| + |AC|$ with equality iff $A = D$, as required.

This result immediately gives that the perimeter of $\triangle ABC$ is greater than the perimeter of $\triangle DBC$, i.e. of all triangles on $BC$ as base inscribed in the circle, the isosceles triangle has the greatest perimeter.

Now suppose the triangle with the greatest perimeter inscribed in a circle is not equilateral. Let $\triangle XYZ$ be the triangle with the greatest perimeter and assume, without loss of generality $|XZ| \neq |YZ|$. From the inequality shown above with $B = X, C = Y$ it follows that $\triangle XYZ$ cannot have the greatest perimeter among all triangles inscribed in the circumcircle of $\triangle XYZ$. This contradiction shows that the triangle with the greatest perimeter that can be inscribed in a circle is the equilateral triangle.