Question

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 1

Given, AB is a chord of a circle with centre O.

AOC is a diameter of the circle

AT is the tangent at point A

We have to prove that ∠BAT = ∠ACB

We know that angle in a semicircle is always equal to 90°

So, ∠CBA = 90°

Considering triangle CBA,

We know that the sum of all three interior angles of a triangle is always equal to 180°

∠CBA + ∠BAC + ∠ACB = 180°

90° + ∠BAC + ∠ACB = 180°

∠BAC + ∠ACB = 180° - 90°

∠BAC + ∠ACB = 90°

∠ACB = 90° - ∠BAC   .......(1)

We know that the radius of the circle is perpendicular to the tangent at the point of contact.

So, OA ⟂ AT

∠OAT + ∠CAT = 90°

From the figure,

∠CAT = ∠BAT + ∠BAC

90° = ∠BAT + ∠BAC

∠BAT = 90° - ∠CAB   ........(2)

Comparing (1) and (2),

Since RHS are same

∠ACB = ∠BAT

Therefore, it is proved that ∠ACB = ∠BAT.

Answer 2

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]