All three parts are about Theorem $\displaystyle 9$ and how far its converse can be pushed.Throughout, X and Y are different from A and B, so that the angles \(\displaystyle \angle \mathrm{AXB}\) and \(\displaystyle \angle \mathrm{AYB}\) make sense. The chord AB cuts the circle into two arcs; for points
on the circle, "on the same side of AB" and "on the same arc" mean the same thing.
(i) No — there are no such points.Suppose X and Y are both on the circle and on the same side of AB, i.e. both on the same arc. Call the
other arc, the one neither of them is on, arc AB. Then X and Y are both points of the circle lying outside arc AB, so Theorem $\displaystyle 9$ applies to each of them:
\[\angle \mathrm{AXB}=\tfrac{1}{2}\left(\text{angle arc AB subtends at the centre}\right), \qquad \angle \mathrm{AYB}=\tfrac{1}{2}\left(\text{the same angle}\right). \]
Both are half of one and the same central angle, so they are equal. That is the whole point of Theorem $\displaystyle 9$: the angle does not depend on
which point of that arc you pick (Fig. $\displaystyle 5.23$).
So no such pair X, Y exists.(ii) Not always — it fails exactly when AB is a diameter.Suppose X and Y are on the circle but on
opposite arcs. Let the arc containing Y subtend the angle \(\displaystyle \alpha\) at the centre; then the arc containing X subtends the rest of a full turn, \(\displaystyle 360^\circ-\alpha\). Apply Theorem $\displaystyle 9$ to each point (each is outside the arc containing the other):
\[\angle \mathrm{AXB}=\tfrac{\alpha}{2}, \qquad \angle \mathrm{AYB}=\tfrac{360^\circ-\alpha}{2}=180^\circ-\tfrac{\alpha}{2}. \]
Adding,
\[\angle \mathrm{AXB}+\angle \mathrm{AYB}=180^\circ. \]
So points on opposite arcs always see AB at
supplementary angles. Two supplementary angles are equal only when each is \(\displaystyle 90^\circ\), which needs \(\displaystyle \alpha=180^\circ\) — that is, AB is a diameter.
Therefore:
| If AB is not a diameter | \(\displaystyle \angle \mathrm{AXB}=\angle \mathrm{AYB}\) does force X, Y onto the same arc |
| If AB is a diameter | every point of the circle sees AB at \(\displaystyle 90^\circ\) |
The second row is the corollary to Theorem $\displaystyle 9$: the angle subtended by a diameter at any point of the circle is \(\displaystyle 90^\circ\). So if AB is a diameter, take X above AB and Y below it — then \(\displaystyle \angle \mathrm{AXB}=\angle \mathrm{AYB}=90^\circ\) although X and Y are on opposite sides.
The statement as written is therefore not true in general.(iii) Yes if X and Y are on the same side of the line AB; not otherwise.Same side. A, B, X are not collinear, so by Theorem $\displaystyle 1$ there is exactly one circle through A, B and X; draw it. Suppose Y were not on it. Y is on the same side of AB as X, so Y is either inside or outside this new circle.
If Y is outside, the segment AY crosses the circle at a point E. Then \(\displaystyle \angle \mathrm{AEB}\) is an exterior angle of \(\displaystyle \triangle \mathrm{BEY}\), so \(\displaystyle \angle \mathrm{AEB}>\angle \mathrm{AYB}\). But E and X lie on the same arc of the new circle, so by Theorem $\displaystyle 9$ \(\displaystyle \angle \mathrm{AEB}=\angle \mathrm{AXB}\). Combining with the hypothesis \(\displaystyle \angle \mathrm{AXB}=\angle \mathrm{AYB}\) gives \(\displaystyle \angle \mathrm{AYB}>\angle \mathrm{AYB}\) — impossible.
If Y is inside, extend AY to meet the circle at E. Now it is \(\displaystyle \angle \mathrm{AYB}\) that is the exterior angle of \(\displaystyle \triangle \mathrm{BEY}\), so \(\displaystyle \angle \mathrm{AYB}>\angle \mathrm{AEB}=\angle \mathrm{AXB}\) — again impossible.
Both alternatives are ruled out, so Y lies on the circle through A, B, X. This is exactly Theorem $\displaystyle 10$, which the book proves in the very next section, Section 5.8.
Opposite sides. Then the circle through A, B, X need not pass through Y. For instance, suppose \(\displaystyle \angle \mathrm{AXB}=\angle \mathrm{AYB}=60^\circ\), with X above the line AB and Y below it. Draw the circle through A, B, X. Since \(\displaystyle \angle \mathrm{AXB}=60^\circ\), Theorem $\displaystyle 9$ says the lower arc AB (the one not containing X) subtends \(\displaystyle 2\times 60^\circ=120^\circ\) at the centre, so the upper arc subtends \(\displaystyle 360^\circ-120^\circ=240^\circ\). Applying Theorem $\displaystyle 9$ the other way round, a point of the
lower arc sees AB at \(\displaystyle \tfrac{1}{2}\times 240^\circ=120^\circ\). Y is below AB but sees AB at only \(\displaystyle 60^\circ\), so Y is not one of those points: it is not on the circle. (Part (ii) is the same phenomenon — the two sides give supplementary angles, and \(\displaystyle 60^\circ\) and \(\displaystyle 120^\circ\) are the supplementary pair here.)
Summary: (i) No, by Theorem $\displaystyle 9$ all points of one arc give the same angle. (ii) No, not always — it fails precisely when AB is a diameter, since then every point of the circle sees AB at \(\displaystyle 90^\circ\). (iii) Yes, provided X and Y are on the same side of AB (that is Theorem $\displaystyle 10$); if they are on opposite sides, the circle through A, B, X need not pass through Y. 
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