Why is the angle in a semicircle always 90 degrees?
My textbook states this as a theorem but does not prove it clearly. I want to understand the actual proof, not just memorize the statement.
1 Answer
IMIsha Malhotra
✓ Accepted
· 14d ago
▲ 10
Proof using the angle at centre theorem: Let AB be the diameter of a circle with centre O, and let C be any point on the semicircle. The angle AOB at the centre = 180 degrees (straight line). The theorem states: angle at centre = 2 * angle at circumference. So angle ACB = 180/2 = 90 degrees. Alternatively: OA = OB = OC (all radii). So triangle OAC is isosceles (OA=OC), making angle OAC = angle OCA = alpha. Similarly triangle OBC gives angle OBC = angle OCB = beta. In triangle ABC: alpha + (alpha + beta) + beta = 180, giving 2(alpha + beta) = 180, so angle ACB = alpha + beta = 90 degrees.
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