Q. 44.5( 24 Votes )

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.


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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