In figure, O is the centre of the circle, BD = OD and CD ⊥ AB. Find ∠CAB.
Given, in the figure $B D=O D, C D \perp A B$.
$\ln \triangle O B D$ $B D=O D$ [given]
$O D=O B$ [both are the radius of circle]
$\therefore$ $O B=O D=B D$
Thus, $\triangle O D B$ is an equilateral triangle.
$\therefore \quad \angle B O D=\angle O B D=\angle O D B=60^{\circ}$
In $\triangle M B C$ and $\triangle M B D$, $M B=M B$ [common side]
$\angle C M B=\angle B M D=90^{\circ}$
and $C M=M D$
[in a circle, any perpendicular drawn on a chord also bisects the chord]
$\therefore$ $\triangle M B C \cong \triangle M B D$ [by SAS congruence rule)
$\therefore$ $\angle M B C=\angle M B D$ [by CPCT]
$\Rightarrow$ $\angle M B C=\angle O B D=60^{\circ}$ $\left[\because \angle O B D=60^{\circ}\right]$
Since, $A B$ is a diameter of the circle.
$\therefore \quad \angle A C B=90^{\circ}$
$\begin{array}{ll}\text { In } \triangle A C B, & \angle C A B+\angle C B A+\angle A C B=180^{\circ}\end{array}$
[by angle sum property of a triangle]
$\Rightarrow \quad \angle C A B+60^{\circ}+90^{\circ}=180^{\circ}$
$\Rightarrow \quad \angle C A B=180^{\circ}-\left(60^{\circ}+90^{\circ}\right)=30^{\circ}$
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