Let equal chords AB and CD intersect at P. Since the chords are equal, their corresponding parts have equal sums. Using the intersecting-chords theorem and equality of the chords, we get AP = CP and BP = DP. Hence proved.
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Draw the perpendicular bisectors of any two sides of the cyclic quadrilateral. Their intersection gives the circumcentre. Then observe whether this point lies inside or outside the quadrilateral.
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°.
The angle subtended by an arc at the centre is twice the angle subtended by the same arc at any point on the circle. Therefore, the required angle is 35°.
The perpendicular from the centre to the chord bisects the chord. Using the right triangle, half the chord is 12 cm. Therefore, the complete chord is 24 cm.