In the figure, the diameter passes through O and A lies on the semicircle. Since OA, OB and OC are radii, the relevant triangles are isosceles. Their angles show that a + b = 90°, so the angle at A is 90°.
How would you use the following figure to justify the statement that the angle in a semicircle is 90°?
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Let the endpoints of the diameter be B and C, and let A be a point on the semicircle. Join OA. Since OA, OB and OC are radii, OA = OB = OC. Therefore, the two triangles formed with OA are isosceles. If the angles at B and C are a and b, the corresponding central angles are 180° – 2a and 180° – 2b. Since BOC is a straight angle, a + b = 90°. Hence, ∠BAC = 90°.
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