В равнобедренном треугольнике ABC (AB=AC), в кото...

Finalizing the proof for angle PQC

The angle PQC appears constant, 46°, which matches half of the apex angle (46°). I confirmed that p is within range, so the angle is independent of p. To prove this, I used the symmetry of the isosceles triangle, showed that point P lies on the perpendicular bisector, and demonstrated that the circle centered at P passes through both points B and C.

Understanding the geometry for angle PQC

Now, Q lies on the circle such that PQ = PC, meaning points B, C, and Q all lie on the same circle with center P. Therefore, P is the circumcenter of triangle BCQ. Since triangle PQC is isosceles with PQ = PC, the angle at Q (∠PQC) is the base angle. We're trying to relate this to the triangle geometry.

Exploring angle PQC and geometric relationships

I’m thinking of using circle ω centered at P, passing through B and C, with Q on it. Since AD is the perpendicular bisector of chord BC and AD passes through P, I’m looking at the relationship between the angles. It seems ∠PQC equals half of ∠CAB, which is 46°. Now, to prove it.

Reflecting on circle symmetry and angles

By reflecting across line AD, B and C are symmetric, and the circle centered at P is symmetric as well. Since Q lies on AB such that PQ = PB = PC, it must lie on the circle passing through B and C. This makes triangle BCQ inscribed in that circle. Next, I'll explore the relationship between the radius and chord angles, possibly applying the tangent theorem.

Confirming angle PQC

By applying the tangent-chord theorem, we find that the angle between the radius QP and chord QC equals 90° - the angle between the tangent at Q and QC. Since the inscribed angle subtending arc QC is 44°, angle PQC is 46°. This confirms the final answer of 46°.

Thought for 1m 32s