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 ...
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In a cyclic quadrilateral, opposite interior angles are supplementary. Since an exterior angle and its adjacent interior angle are also supplementary, the exterior angle equals the opposite interior angle. Hence, ∠CDE = ∠ABC.
Each side of a regular hexagon subtends 60° at the centre. The triangle formed is equilateral, so each side is r. The distance from the centre to each side is (√3/2)r.
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.