Question

In the given figure, O is the centre of the circle, BD = OD and CD ⊥ AB. Find ∠CAB.

Answer 1

In the given figure, BD = OD and CD ⊥ AB.



Join AC and OC.

In ∆ODE and ∆DBE,
∠DOE = ∠DBE (given)
∠DEO = ∠DEB = 90^∘
OD = DB (given)
∴ By AAS conguence rule, ∆ODE ≌ ∆BDE,

Thus, OE = EB ...(1)

Now, in ∆COE and ∆CBE,
CE = CE (common)
∠CEO = ∠CEB = 90^∘
OE = EB (from (1))
∴ By SAS conguence rule, ∆COE ≌ ∆CBE,

Thus, CO = CB ...(2)

Also, CO = OB = OA (radius of the circle) ...(3)

From (2) and (3),
CO = CB = OB
∴ ∆COB is equilateral triangle.
∴ ∠COB = 60^∘ ...(4)

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.

Here, arc CB subtends ∠COB at the centre and ∠CAB at A on the circle.

∴ ∠COB = 2∠CAB
⇒ $\angle CAB=\frac{60°}{2}=30°$ (from (4))

Hence, ∠CAB = 30^∘.