
In ∆OAB,
AB = OA = OB = radius
∠∆OAB is an equilateral triangle.
Therefore, each interior angle of this triangle will be of 60°.
∠AOB = 60°
∠ACB=\(\frac{1}{2}\)∠AOB=\(\frac{1}{2}\)(60°)=30°
In cyclic quadrilateral ACBD,
∠ACB + ∠ADB = 180° (Opposite angle in cyclic quadrilateral)
∠ADB = 180° − 30° = 150°
Therefore, angle subtended by this chord at a point on the major arc and the minor arc are 30° and 150° respectively.
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig. 9.27). Prove that ∠ACP = ∠ QCD


(i) The kind of person the doctor is (money, possessions)
(ii) The kind of person he wants to be (appearance, ambition)
ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see Fig. 7.17). Prove that
(i) ∆ ABD ≅ ∆ BAC
(ii) BD = AC
(iii) ∠ ABD = ∠ BAC.
