Irska

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 $$\begin{array}{rl}|DB|+|DC|& =2r\cos \angle BDO+2r\cos \angle CDO\\ & =2r(\cos \angle BDO+\cos \angle CDO)\\ & =4r\cos \left(\frac{\angle BDO+\angle CDO}{2}\right)\cos \left(\frac{\angle BDO-\angle CDO}{2}\right)\end{array}$$ 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 $△ABC$ is greater than the perimeter of $△DBC$, 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 $△XYZ$ be the triangle with the greatest perimeter and assume, without loss of generality $|XZ|\ne |YZ|$. From the inequality shown above with $B=X,C=Y$ it follows that $△XYZ$ cannot have the greatest perimeter among all triangles inscribed in the circumcircle of $△XYZ$. This contradiction shows that the triangle with the greatest perimeter that can be inscribed in a circle is the equilateral triangle.