In a triangle ΔPQR, ∠R = 62° . The perpendicular bisector of PQ at S meets QR at T. If ∠TPR = 38° , what is the measure of ∠PQR?

Correct — D. T lies on the perpendicular bisector of PQ, so it is equally far from P and Q.

Equal distances: TP = TQ, so triangle TPQ is isosceles
Equal base angles: ∠TPQ = ∠TQP
T lies on QR, so ∠TQP is ∠PQR itself; call it x

Split ∠P at T: ∠QPR = ∠QPT + ∠TPR = x + 38°
Angle sum of PQR: (x + 38°) + x + 62° = 180°
Collect: 2x + 100° = 180°, so 2x = 80°
Halve: x = 40° → option (d)

Why the others are wrong

• (a)
  60° — Test it: with ∠Q = 60°, ∠P = 60° + 38° = 98°, and 98° + 60° + 62° = 220°. The angles of triangle PQR must total 180°.
• (b)
  80° — 80° is 2x, the two equal base angles together: 180° − 62° − 38° = 80°. Halve it to get ∠PQR = 40°.
• (c)
  35° — Test it: with ∠Q = 35°, ∠P = 35° + 38° = 73°, and 73° + 35° + 62° = 170°, which falls 10° short of 180°.

Every point on the perpendicular bisector of a segment is equidistant from its two ends. So T, sitting on the bisector of PQ, makes TP = TQ.

Equal sides give equal base angles. That copies the unknown angle at Q across to P, and the angle sum of the big triangle finishes the job.

Point S, the foot of the bisector on PQ, plays no part in the angle chase.

Key facts

• Any point on the perpendicular bisector of PQ is equidistant from P and Q.
• In an isosceles triangle, the angles opposite the equal sides are equal.
• Here ∠PQR = 40°, ∠QPR = 78° and ∠PRQ = 62°, which total 180°.

Common traps

• Stopping at 180° − 62° − 38° = 80° and forgetting that this is two equal angles, not one.
• Taking ∠TPR = 38° as the whole of ∠P instead of adding ∠TPQ to it.

No directly related past PYQ was found.