A quadrilateral ABCD is called cyclic if all the four vertices of it lie on a circle in fig. A quadrilateral is said to be a cyclic quadrilateral, if there is a circle passing through all its four vertices.
`=>`∠A + ∠C = 180°
also ∠B + ∠D = 180°.
Theorem: The sum of either pair of opposite angles of a cyclic quadrilateral is 180º.
Given : A cyclic quadrilateral ABCD .
To Prove : ∠BAD + ∠BCD = ∠ABC + ∠ADC = 180º.
Construction : Draw AC and DB.
Proof : ∠ACB = ∠ ADB and
∠BAC = ∠BDC [Angles in the same segment]
`therefore` ∠ACB + ∠BAC = ∠ADB +∠BDC = ∠ADC
Adding ∠ABC on both the sides, we get
∠ACB + ∠BAC +∠ABC = ∠ADC +∠ABC
But ∠ACB +∠BAC +∠ABC = 180º. [sum of the angles of a triangle]
`therefore` ∠ ADC + ∠ABC =180º.
`therefore` ∠BAD +∠BCD = 360º - (∠ADC + ∠ABC) = 180º.
Theorem: If the sum of a pair of opposite angles of a quadrilateral is 180º, the quadrilateral is cyclic.
Shaalaa.com | Cyclic Quadrilaterals
In the given figure, A, B, C and D are four points on a circle. AC and BD intersect at a point E such that ∠BEC = 130° and ∠ECD = 20°. Find ∠BAC.
AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters; (ii) ABCD is a rectangle.
If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
Two chords AB and CD of lengths 5 cm 11cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.