For any chord through A, its length depends on its distance from O. The chord is shortest when this distance is greatest. This occurs when the chord is perpendicular to OA. Hence proved.
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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 ...
Since AB and AC are congruent chords, they are equidistant from O. In triangles AOB and AOC, OA is common and OB = OC. Hence, the triangles are congruent, giving ∠BAO = ∠OAC. Therefore, AO bisects ∠BAC.
Let ABCD be a parallelogram inscribed in a circle. Opposite angles of a cyclic quadrilateral are supplementary, while opposite angles of a parallelogram are equal. Therefore, each angle is 90°. Hence, ABCD is a rectangle.
Since AB is the diameter of the circle, the angle subtended by the diameter at any point on the circumference is 90°. Therefore, ∠ACB = 90°.