AB is a chord of the circle with centre O AOC is the diameter and AT is the tangent at A. Proove that angle BAT = angleACB (use alternate angle to proove)

Sure, here is the step by step proof:

Step 1: We know that the line from the center of a circle perpendicular to a chord bisects the chord. So, OC bisects AB at point D.

Step 2: Now, we have two triangles, ∆OAD and ∆OCD. In these triangles, we have OA = OC (radii of the same circle), OD = OD (common), and AD = CD (from step 1).

Step 3: Therefore, by the Side-Side-Side (SSS) criterion, ∆OAD ≅ ∆OCD.

Step 4: So, ∠OAD = ∠OCD (corresponding angles of congruent triangles are equal).

Step 5: We also know that the angle between a tangent and a radius through the point of contact is 90 degrees. So, ∠OAT = 90 degrees.

Step 6: Therefore, ∠BAT = ∠OAT - ∠OAD = 90 - ∠OAD.

Step 7: Similarly, ∠ACB = ∠OCD - ∠OCA = ∠OAD - 90 (since ∠OCA is a straight line and is 180 degrees, and ∠OCD + ∠OCA + ∠ACB = 180).

Step 8: From steps 6 and 7, we can see that ∠BAT = ∠ACB.

So, we have proved that ∠BAT = ∠ACB using alternate angles.