Basic Properties of Ciricles

Since $AB$ is a diameter of the circle, $\angle AXB =90^\circ$ . Since $ABYX$ is a cyclic quadrilateral , $\angle BAY = \angle BXY = \theta$ . Then for $\triangle AXY$ we have:

$\begin{aligned} \angle AXZ + \angle XAZ + \angle XZA & = 180^\circ \\ 90^\circ + \theta + 31^\circ + \theta + 21^\circ & = 180^\circ \\ 2 \theta & = 38^\circ \\ \implies \theta & = 19^\circ \end{aligned}$

And for $\triangle ABX$ we have $\angle ABX = 180^\circ - \angle BXA - \angle XAB = 180^\circ - 90^\circ - (31^\circ + 19^\circ) = \boxed{40}^\circ$ .

Let $\angle {BAY}=\angle {BXY}=α$ . Then $\angle {ABX}=21\degree+α$ . Since $\angle {AXB}=90\degree$ , therefore $\angle {XAB}+\angle {ABX}=90\degree\implies 31\degree+α+21\degree+α=90\degree\implies α=19\degree\implies \angle {ABX}=21\degree+α=\boxed {40\degree}$ .