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 Fig. 9.17. Prove that ∠BAT = ∠ACB

Home » 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 Fig. 9.17. Prove that ∠BAT = ∠ACB

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 Fig. 9.17. Prove that ∠BAT = ∠ACB

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 Fig. 9.17. Prove that ∠BAT = ∠ACB

As per the inquiry,

A circle with focus O and AC as a measurement and AB and BC as two harmonies additionally AT is a digression at point A

To Prove : ∠BAT = ∠ACB

Verification :

∠ABC = 90° [Angle in a half circle is a right angle]

In △ABC By point total property of triangle

$\angle ABC\text{ }+\angle BAC\text{ }+\angle ACB\text{ }=\text{ }180{}^\circ$

$\angle ACB\text{ }+\text{ }90{}^\circ =\text{ }180{}^\circ \angle BAC$

$\angle ACB\text{ }=\text{ }90\angle BAC\text{ }\left[ 1 \right]$

Presently,

OA ⏊ AT [Tangent at a point on the circle is opposite to the sweep through resource ]

$\angle OAT\text{ }=\angle CAT\text{ }=\text{ }90{}^\circ$

$\angle BAC\text{ }+\angle BAT\text{ }=\text{ }90{}^\circ$

$\angle BAT\text{ }=\text{ }90{}^\circ \angle BAC\text{ }\left[ 2 \right]$

From [1] and [2]

∠BAT = ∠ACB [Proved]

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