Ac is a diameter of circle centre O. B is a point on the circle. OP bisects AB. Prove that OP||BC
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.
To prove that OP is parallel to BC, we need to show that the opposite interior angles formed by these lines are equal.
Given that AC is a diameter and OP bisects AB, we know that angle AOB is a right angle. This is because a diameter of a circle is always perpendicular to any chord that it bisects.
Let's label the points for clarity:
- Let O be the center of the circle.
- Let A be the point where AC intersects the circle.
- Let B be a point on the circle.
- Let P be the midpoint of AB.
Since OP is the perpendicular bisector of AB, we can conclude that angle OPA is also a right angle. This is because the perpendicular bisector of a line segment creates right angles with the line segment at the point of intersection.
Now, let's consider the triangle OAB. From the above discussion, we have:
Angle AOB = 90 degrees (given)
Angle OPA = 90 degrees (proved)
Since angles OPA and AOB are both 90 degrees, they are equal.
By the Converse of the Alternate Interior Angles Theorem, if two lines intersected by a transversal have equal alternate interior angles, then the lines are parallel. In this case, the transversal is BC, and the alternate interior angles are angle AOB and angle OPA.
Therefore, OP is parallel to BC, as required.