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Question 14
In the figure, O is the centre of the circle BD = OD and CD AB. Find CAB.

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Solution

Given, in the figure BD = OD, CD AB

In Δ OBD,
BD = OD
OD = OB [both are the radius of same circle]
OB = OD = BD

Thus, Δ ODB is an equilateral triangle.
BOD=OBD=ODB=60

In Δ MBC and Δ MBD
MB = MB [common]
CMB=BMD=90
And CM = MD [in a circle, any perpendicular drawn on a chord also bisect the chord]
ΔMBCΔMBD [ by SAS congruence]
MBC=MBD [by CPCT]
MBC=OBD=60
Since, AB is a diameter of the circle
ACB=90

In ΔACB CAB+CBA+ACB=180 [by angle sum property of a triangle]
CAB+60+90=180
CAB=180(60+90)=30


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