In Fig. 10.21, O is the centre of the circle, BD = OD and CD AB. Find CAB.
30°
Solution:
Given: In the figure BD = OD, CD AB
In OBD,
BD = OD (given)
OD = OB (radius of the same circle)
OB = OD = BD
Hence the triangle is equilateral
BOD = OBD = ODB = 60°
Consider MBC and MBD
MB = MB (common)
CMB = BMD = 90° (given)
CM = MD (perpendicular from the centre on the chord bisects the chord]
MBC =MBD (SAS rule)
MBC =MBD (CPCT)
MBC = OBD = 60° (OBD = 60°)
Since AB is the diameter of the circle
ACB = 90° (angle is a semi-circle)
CAB +CBA + ACB = 180° (By angle sum property of a triangle)
Putting the values,
CAB + 60° + 90° = 180°
CAB = 180° – (60° + 90°) = 30°