Solving angles within a cyclic quadrilateral

Question

Please could you help with solving angles x and y as well as writing how you solved them.

Answer 1

The Inscribed Angle Theorem says that the angle subtended at the centre by a chord is twice the angle subtended at (the longer arc of) the circumference. This gives you x immediately, from chord $AB$.

Now you know $\angle CDA$, because $OB$ and $CD$ are parallel. This gives you $\angle ABC$, because opposite angles of a cyclic quadrilateral add up to $180^{\circ}$.

And now you have two angles of triangle $ABC$, so you also know the third angle y.

Answer 2

Hint: $ACD$ is a right triangle since $AD$ is a diameter. Further, $AEO$ is also a right triangle since $OB$ and $CD$ are parallel. From here, we conclude that $BO$ bisects chord $AC$ since it is a (piece of a) diameter perpendicular to it. Now think congruent triangles.

Another hint: The interior angles of a triangle add to $180$ degrees, and those of a quadrilateral add to $360$ degrees.