Q20 of 100 Page 17

Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.

Let O be the circle circumscribing the cyclic rectangle ABCD.

Since, ∠ABC = 90^o and AC is the chord of the circle. Similarly, BD is a diameter

Hence, point of intersection of AC and BD is the centre of the circle.

ABCD is a cyclic quadrilateral in which:

(i) BC||AD, ∠ADC=110° and ∠BAC=50°. Find ∠DAC.

(ii) ∠DBC=80° and ∠BAC=40° Find ∠BCD.

(iii) ∠BCD=100° and ∠ABD=70°. Find ∠ADB.

Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.

Prove that the circles described on the four sides of a rhombus as diameters, pass through the point of intersection of its diagonals.

If the two sides of a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are equal.