Question

AB is a chord of a circle with centre O, AOC is a diameter and AT is the tangent at A as shown in Figure. Prove that $\angle BAT=\angle ACB$.

Answer 1

Given: A circle with center O and AC as a diameter and AB and BC as two chords also AT is a tangent at a point A​

The angle in a semi-circle is right angle,

so $\angle ABC={90}^0$.

$\angle ABC+\angle BAC+\angle ACB=180$ (Angle sum property of a triangle)

Therefore, $90+\angle ACB=180-\angle BAC$

$\Rightarrow \angle ACB=90-\angle BAC$............(1)
Now OA perpendicular AT
Therefore, $\angle OAT=\angle CAT=90$

$\angle BAC+\angle BAT=90$

$\angle BAT=90-\angle BAC$.....(2)

Hence ,from (1) and (2)

$\angle BAT=\angle ACB.$

Hence, proved.