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Question
If AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in figure. Prove that ∠BAT = ∠ACB
Solution
Since, AC is a diameter line, so angle in semi-circle an angle 90°.
∴ ∠ABC = 90° ...[By property]
In ∆ABC,
∠CAB + ∠ABC + ∠ACB = 180° ...[∵ Sum of all interior angles of any triangle is 180°]
⇒ ∠CAB + ∠ACB = 180° – 90° = 90° ...(i)
Since, diameter of a circle is perpendicular to the tangent.
i.e., CA ⊥ AT
∴ ∠CAT = 90°
⇒ ∠CAB + ∠BAT = 90° ...(ii)
From equations (i) and (ii),
∠CAB + ∠ACB = ∠CAB + ∠BAT
⇒ ∠ACB = ∠BAT
Hence proved.
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