Ac is a diameter of circle centre O. B is a point on the circle. OP bisects AB. Prove that OP||BC

Answer 1

Since OP bisects AB, we have AP = PB. Since AC is a diameter, ∠ABC = 90°.

Now consider triangles ABC and OBC. They share side BC and have a common vertex at B.

By the Angle Bisector Theorem, we have:

$\frac{AP}{PB}=\frac{AO}{OB}$

Since AP = PB, we can rewrite this as:

1 = $\frac{AO}{OB}$

This means that AO = OB, and so triangle OBC is isosceles. Therefore, ∠OBC = ∠OCB.

But since ∠ABC = 90°, we also have:

∠OBC + ∠OCB = 90°

Combining these two equations, we get:

2∠OBC = 90°

∠OBC = 45°

Now consider triangles OBP and OPC. They share side OP and have a common vertex at O.

Since we just showed that ∠OBC = 45°, we have:

∠OPC = 180° - ∠OBC = 180° - 45° = 135°

But since the sum of angles in a triangle is 180°, we have:

∠OPB + ∠OPC = 180° - ∠BOP = 180° - 90° = 90°

Therefore, ∠OPB = 45°.

Since ∠OPB = ∠OBC, we have OP || BC by the Converse of Alternate Interior Angles Theorem.

Therefore, we have proved that OP || BC.