S is a point on side PQ of a triangle PQR such that RS = PS = QS. If ∠ QPR =…

Concept

A point that is equidistant from all three vertices of a triangle is its circumcentre, and it is the centre of the circle passing through the three vertices. If that point lies on one side of the triangle, then that side is a diameter of the circle, so the angle subtended at the opposite vertex is a right angle (the angle in a semicircle, Thales' theorem). Two further facts are used: the base angles of an isosceles triangle are equal, and the three interior angles of a triangle sum to 180°.

Application

Here S lies on side PQ and RS = PS = QS, so S is equidistant from P, Q and R. Working through it:

1. Since PS = QS = RS, the point S is the centre of a circle passing through P, Q and R, and this centre lies on side PQ.

2. Because S is on PQ with PS = QS, the chord PQ passes through the centre, so PQ is a diameter of that circle.

3. R lies on the circle, so ∠PRQ is the angle in a semicircle and therefore ∠PRQ = 90°.

4. In triangle PQR the angles sum to 180°, so ∠PQR = 180° − ∠QPR − ∠PRQ = 180° − 15° − 90° = 75°.

5. Hence 2 ∠PQR − ∠PRQ = 2 × 75° − 90° = 150° − 90° = 60°.

Cross-check

The same result follows from the isosceles triangles directly. In triangle PSR, PS = RS gives ∠PRS = ∠QPR = 15°. In triangle QSR, QS = RS gives ∠QRS = ∠PQR. Since ∠PSR and ∠QSR are supplementary, the exterior-angle relation forces ∠QRS = 75°, so ∠PRQ = ∠PRS + ∠QRS = 15° + 75° = 90°, matching the semicircle argument and again giving 2 × 75° − 90° = 60°.