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If AB is a chord of a circle with center O, AOC is a diameter and AT is the tangent at A as shown in figure. Prove that ∠BAT = ∠ACB. |
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Answer» Given : A circle with center O and AC as a diameter and AB and BC as two chords also AT is a tangent at point A To Prove : ∠BAT = ∠ACB Proof : ∠ABC = 90° [Angle in a semicircle is a right angle] In △ABC By angle sum property of triangle ∠ABC + ∠ BAC + ∠ACB = 180 ° ∠ACB + 90° = 180° - ∠BAC ∠ACB = 90 - ∠BAC [1] Now, OA ⏊ AT [Tangent at a point on the circle is perpendicular to the radius through point of contact ] ∠OAT = ∠CAT = 90° ∠BAC + ∠BAT = 90° ∠BAT = 90° - ∠BAC [2] From [1] and [2] ∠BAT = ∠ACB [Proved] |
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