As per the question, the answer is ∠CAB is congruent to ∠BAD.
An isosceles triangle named BDC is positioned within a circle. The segments of the circle they pass through are congruent because the faces of the triangles BC and BD are congruent, or identical.
Due to the triangle's faces BC and BD being identical and congruent
BC=BD
Additionally, the segments of the circle it goes across are congruent.
In the case of ΔBAC & ΔBAD
AC=AD= the radius of the circle
AB=BA
Thus, BAC is consistent with BAD according to SSS guidelines.
As a result, the central angles of the sections are also congruent; hence, BAD and CAB are congruent.
As a result, ∠BAD is equivalent to ∠BAC OR ∠CAB.
Learn more about SSS guidelines:
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