Question

Triangle ABC is inscribed in a circle with centre O, AB = AC, OP ⊥ AB and OQ ⊥ AC. Prove that PB = QC.

Answer 1

Given:

• Triangle ABC is inscribed in a circle with center O.
• AB = AC triangle ABC is isosceles.
• OP ⊥ AB and OQ ⊥ AC the perpendiculars from the center of the circle to the sides AB and AC.

To Prove:

We need to prove that PB = QC, where P is the foot of the perpendicular from O to AB and Q is the foot of the perpendicular from O to AC.

Proof:

1. Isosceles Triangle: Since AB = AC, triangle ABC is isosceles. Therefore, the perpendicular from O to AB and the perpendicular from O to AC will have special symmetry.

2. Perpendicular Bisectors: The perpendiculars OP and OQ are both drawn from the center O of the circle. Since AB = AC, the perpendiculars are equally spaced from the center and divide the triangle symmetrically.

3. Congruent Right Triangles: We can now focus on the two right triangles formed by these perpendiculars.

Specifically, we look at the two right triangles OPB and OQC, where:

• OP ⊥ AB
• OQ ⊥ AC
• OP = OQ   ...(Since both are radii of the circle)
• ∠OPB = ∠OQC = 90°   ...(Right angles)

By the hypotenuse-leg (HL) congruence criterion for right triangles, since OP = OQ and OB = OC are radi of the same circle and ∠OPB = ∠OQC = 90°, we conclude that ΔOPB ≅ ΔOQC

4. Corresponding Parts of Congruent Triangles: Since the triangles ΔOPB and ΔOQC are congruent, the corresponding parts of these triangles must be equal.

In particular: PB = QC

Thus, we have proved that PB = QC as required.