BMO Dec 2002 Q2: Circles and triangles. Steps showing construction would be greatly appreciated

Question

The triangle ABC, where AB < AC, has circumcircle S. The perpendicular from A to BC meets S again at P. The point X lies on the segment AC and BX meets S again at Q.

Show that BX = CX if and only if PQ is a diameter of S.

I

What is really throwing me off is that I don't know how to construct this diagram such that PQ is a diameter, especially because I don't know if it matters how the triangle is done. I know the triangle can't be equilateral cuz AB < AC but I don't know if it matters whether it is an isosceles triangle or a scalene triangle. Furthermore, I thought about constructing the diagram about PQ first but that didn't work out for me. Also, I have included 2 images (source material below in case they don't work, for the book go to page 72) but I have no idea what is going on in either of them. I want to know the reasoning cuz I've already go the answer but I'm open to other methods as long as they are explained well (step by step, preferably including geometric properties used). Thanks in advance.

https://bmos.ukmt.org.uk/home/bmomarking.pdf https://www.google.com/books/edition/Lecture_Notes_On_Mathematical_Olympiad_C/5n5IDQAAQBAJ?hl=en&gbpv=1

Answer 1

1. Draw a circle S with center O.

2. Let A, B, C be any three points on the circle.

3. Extend the perpendicular from A to CD and let it cut S at P.

4. Extend PO to cut S again at Q.

5. Let BQ cut AC at X.

Answer 2

$$\angle CBP + \angle BCA = \angle CAP + \angle BCA = 90^\circ$$

The first equality is due to $\angle CBP = \angle CAP$ (angles in the same segment).

The second equality is due to $AP \perp BC$, and that the sum of interior angles of a triangle is $180^\circ$.

$$\angle CBP + \angle QBC = \angle QBP$$

is obvious by inspection.

$$\therefore PQ \text{ is a diameter of }S \iff \angle QBP = 90^\circ$$

This is Thales's theorem and its converse.

$$\iff \angle CBP + \angle QBC = 90^\circ$$ $$\iff \angle BCA = \angle QBC$$

These are obvious by manipulating our previous results.

$$\iff BX=CX$$

This is the determination and property of an isosceles triangle.