Let A, B, C be three points on a circle of centre O. The perpendicular line from O to BC intersects line AC at P, and the perpendicular line from O to AC intersects line BC at Q. Let L be the midpoint of OC and K the midpoint of PQ. Prove that KL is perpendicular on AB.
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The quadrilateral CNOM is cyclic because ∠CNO=90∘ and ∠CMO=90∘, and OC is a diameter and L the centre of its circumcircle. The quadrilateral MNPQ is cyclic as well since ∠PNQ=∠PMQ=90∘, and PQ is a diameter and K the centre of its circumcircle. The line MN is the radical axis of these two circles and so is perpendicular to the line KL which connects the centres of the circles. Because M and N are midpoints of sides of △ABC, MN∥AB, hence AB⊥KL.
Source: MathNet,
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