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

To prove: Perpendicular bisector of side AB, BC, CD and DA are concurrent i.e, passes through the same point.
Proof:
We know that the perpendicular bisector of every chord of a circle always passes through the centre.
Therefore, Perpendicular bisectors of chord AB, BC, CD and DA pass through the centre which means they all passes through the same point.
Hence, the perpendicular bisector of AB, BC, CD and DA are concurrent.
APPEARS IN
RELATED QUESTIONS
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
In a cyclic quadrilateral ABCD, if ∠A − ∠C = 60°, prove that the smaller of two is 60°
ABCD is a cyclic quadrilateral in BC || AD, ∠ADC = 110° and ∠BAC = 50°. Find ∠DAC.
Prove that the circles described on the four sides of a rhombus as diameters, pass through the point of intersection of its diagonals.
ABCD is a cyclic trapezium with AD || BC. If ∠B = 70°, determine other three angles of the trapezium.
ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that EB = EC.
In the given figure, ABCD is a cyclic quadrilateral in which ∠BAD = 75°, ∠ABD = 58° and ∠ADC = 77°, AC and BD intersect at P. Then, find ∠DPC.

ABCD is a cyclic quadrilateral such that ∠A = 90°, ∠B = 70°, ∠C = 95° and ∠D = 105°.
If a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are also equal.
