 Hello and welcome to the session. In this session, we discussed the final question that says, in the given figure, A B is the diameter and A C is the part of the circle such that under B A C is equal to 30 degrees, the tangent at C intersects A B produced at B, prove that B C is equal to B D. Before we move on to the solution, let's discuss the alternate segment property of the circle. It is compact, it is drawn, the angle is equal to the angle is the key idea that we use for this question. This is the solution. This is the figure given to us. See what all we have. We are given at A B is the diameter of the circle above the circle and then B A C is of measure 30 degrees. Tangent at C means A B produced at the point D and we are supposed to prove that B C is equal to B D. The point that is conscious of the circle is on the point that is compact, a chord is drawn, the angles between the tangent and the chord are respectively equal to the angles of the alternate that the angle between the tangent C D and the chord B C that is this angle would be equal to this angle. The angle would be equal to C is of measure 30 degrees. So this means angle B C D is equal to 30 degrees. So this is of measure 30 degrees. A C B would be of measure 90 degrees since the angle is of measure 90 degrees and the angle is 90 degrees. B A C we know this means that angle is equal to 30 degrees 90 degrees which is equal to 120 degrees. C B D is of measure 120 degrees. B C D plus is equal to 180 degrees. D which is of measure 30 degrees. B D B. D B C which is of measure 120 degrees is equal to 180 degrees. C D B is equal to 180 degrees, 50 degrees that is 120 degrees plus 30 degrees is not 50 degrees and this means angle C D B is equal to 30 degrees. This is of measure we have B C is equal to B D as we know that B C is equal to B D. So this comes in C session hope you have understood the solution of this question.