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.
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∠MOP is the angle subtended by arc MP at the centre, while ∠MNP is subtended by the same arc at the circumference. Therefore, the central angle is twice the angle at the circumference.
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.
Half of the chords are 5 cm and 12 cm. Let the distance of the 24 cm chord from the centre be x. Then the other is x + 7. Solving gives x = 5 and radius = 13 cm.
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.