It’s kind of a geometrical and symmetrical idea. I am not sure of how to pen down the exact words, but it has something to do with either congruency or symmetry.

For this question we can refer back to the proof of this theorem :


https://proofwiki.org/wiki/Perpendicular_Bisector_of_Chord_Passes_Through_Center


Basically, prove two triangles are common.
Then rationalise that one of their congruent angles is a right angle (i.e the angles made by the intersection of the line drawn from the centre of the circle and the chord)

(sum of angles on a straight line = 180° and the two angles are equal, so each angle = 90°)

Since the perpendicular bisector is drawn, the angles made by the intersection of the chord and the bisector are also right angles.

This means that these angles are common angles with the triangles' right angles.

Therefore the centre of the circle and the bisector are collinear (lie on the same line)
and so the bisector passes through the centre.

You probably won't have to prove it, just need to state that AB is a chord of the circle since the circle passes through both points. Then quote the theorem.

A bit unusual to quote the chord and the theorem just like that for the mark.

It’s like saying, I’m an adult because I’m not a primary school student.

Technically yes though, the bisector presents a form of symmetrical aspects of the circle as the bisector cuts the circle in half.

Well, the question doesn't explicitly ask to prove, and the proofs aren't thoroughly covered at their level either.

It's the same case when other properties are used ( tangent perpendicular to radius, angles in the same segment, etc) Students just need identify the theorem to use and make the relevant statements

Anyway since it's only part a) of the question and comparing to the other parts of the which require more detailed working,

a) is probably worth 2 marks at most so a full explanation with proof would be too much for that.