# 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.

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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