prove that the centre ofthecircle circumscribe the cyclic rectangle ABCD is the point of intersection of its diagonals.
this is for 3 marks
can u plz 1st explain the ques nd then the ans...............
Given, ABCD is a cyclic rectangle whose diagonals intersect at O.
To show : O is the centre of circle.
We know that ∠BCD = 90° (since it is a rectangle)
So, BD is the diameter of the circle (If angle made by the chord at the circle is right angle then the chord is the diameter).
Also, diagonals of a rectangle bisect each other and are equal.
∴ BO = OD
BD is the diameter therefore, BO and OD are the radius.
Thus, O is the centre of the circle.
Hence, centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.