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
![\[\angle ABC\text{ }+\angle BAC\text{ }+\angle ACB\text{ }=\text{ }180{}^\circ \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-d9f901d531cf4f422762a0d8faf20168_l3.png "Rendered by QuickLaTeX.com")
![\[\angle ACB\text{ }+\text{ }90{}^\circ =\text{ }180{}^\circ \angle BAC\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-fbdbf0760d0a5bf5f5fe23e66e692a05_l3.png "Rendered by QuickLaTeX.com")
![\[\angle ACB\text{ }=\text{ }90\angle BAC\text{ }\left[ 1 \right]\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-72c61b6f62350227268cab9c790bd4e6_l3.png "Rendered by QuickLaTeX.com")
Presently,
OA ⏊ AT [Tangent at a point on the circle is opposite to the sweep through resource ]
![\[\angle OAT\text{ }=\angle CAT\text{ }=\text{ }90{}^\circ \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-52ca73f23e1893a72141cef086f840c0_l3.png "Rendered by QuickLaTeX.com")
![\[\angle BAC\text{ }+\angle BAT\text{ }=\text{ }90{}^\circ \]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-561c1d590a384be90345ba6f75de461d_l3.png "Rendered by QuickLaTeX.com")
![\[\angle BAT\text{ }=\text{ }90{}^\circ \angle BAC\text{ }\left[ 2 \right]\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-23b3721e8c6920a9914ce1fa57f82df9_l3.png "Rendered by QuickLaTeX.com")
From [1] and [2]
∠BAT = ∠ACB [Proved]