Let $E$ be the reflection of $A$ across $D$. By Power of a Point, $ABDC$ is cyclic. Now let $E'$ be a point on $(ABDC)$ such that $EE' \parallel BC$ and $D'$ be the intersection of $AE'$ and $BC$. Hence, $DD'$ must be a midline of $AEE'$, $D'$ must be the midpoint of $AE'$, and $D'B \cdot D'C = D'A \cdot D'E = D'A^2$. Since $D$ must be a unique point on $BC$, $E = E'$ which means $AD$ must be the angle bisector of $\angle BAC$. By angle bisector theorem, $\frac{AB}{BD} = \frac{AC}{CD} = k$ for some real $k$. By Stewart's, $BD \cdot CD \cdot BC + AD^2 \cdot BC = AB^2 \cdot DC + AC^2 \cdot BD$ $BD \cdot CD + BD \cdot CD = k^2 \cdot BD^2 \cdot DC + k^2 \cdot CD^2 \cdot BD$ $2 \cdot BD \cdot CD = k^2 \cdot BD \cdot CD \cdot (BC + CD)$ $2 = k^2$ So $k = \sqrt{2}$ and $AB + BC + CA = \sqrt{2} \cdot BD + \sqrt{2} \cdot CD + 1 = \sqrt{2} \cdot (BD + CD) + 1 = \boxed{\sqrt{2} + 1}$.

What motivated you to draw a line $EE' \parallel BC$?

A-Thought-Of-God wrote: What motivated you to draw a line $EE' \parallel BC$? I was trying to look for another point on $BC$ which satisfied the above length condition. Then I thought of midlines since there is a midpoint involved and thus was inspired to draw a parallel line so that $BC$ can be a midline.