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.
Using Brahmagupta’s formula for a cyclic quadrilateral, with sides 5, 5, 12 and 12 units, the semiperimeter is 17 units. Therefore, area = √(12 × 12 × 5 × 5) = 60 square units.
The perpendicular from the centre to the chord bisects it. Half the chord is 8 cm. Using the right triangle, radius² = 8² + 6² = 100. Therefore, radius = 10 cm.
Opposite angles of a cyclic quadrilateral are supplementary. (2x + 10) + (3x – 20) = 180 5x – 10 = 180 x = 38 Thus, ∠P = 86° and ∠R = 94°.