 Hello and welcome to the session. In this session, we discussed the following question which says in a right triangle ABC, a circle with AB as diameter is drawn to intersect the hypotenuse AC in P, prove that the tangent at P by sec C side BC. This is the figure in which we have ABC is a right triangle and the circle with AB as the diameter is drawn such that if intersect C hypotenuse AC at the point P, we have to prove that this tangent at the point P by sec C side BC. First of all, we will discuss the alternate segment property. According to this, we have if straight line touches a circle from the point of the angles between the P equal to the angles of the alternate segments. This is the key idea that we use for this question. Here is how we move on to the solution. This is the figure in which we have is a right triangle such that and the ABC is equal to 90 degrees then at the point P circle is drawn taking ABC as the diameter and it intersects the hypotenuse AC at point P. And we are supposed to prove that the tangent at P by sec C side BC that is we have to prove BQ is equal to QC. First of all, we join PB start with the pose APB would be of measure 90 degrees that is this angle since we know the circle is 90 degrees and this is the angle of the circle. Now, plus BPC is equal to 180 degrees. So these two angles form a linear pair. So this means these are pressure 90 degrees so under BPC is equal to 180 degrees minus 90 degrees which is equal to 90 degrees. That we have BPC is equal to 90 degrees that is this angle is also of measure 90 degrees. Now, from the figure angle BPC is equal to angle BPC is equal to QPC and so this is equal to 90 degrees that is angle BPC plus angle QPC is equal to 90 degrees. Make this result we result 1. Next we consider the triangle ABC in this angle B plus angle C is equal to 180 degrees by B angle some property. So this means C is equal to 180 degrees minus 90 degrees since we know that angle B is equal to 90 degrees as triangle ABC is a right triangle. So here we have angle A plus angle C is equal to 90 degrees like this we result 2. So we have therefore angle between the tangent and the chord that is angle BPC would be equal to the angle in the alternate segment that is this angle which is angle which is using the property. According to which we have that if a straight line touches a circle and from the point of contact a chord is drawn the angles between the tangent and the chord are respectively equal to the angles in the alternate segments. BPC is equal to B same as angle A. Let this be result 3. Now in result 1 we have angle BPC plus angle QPC is equal to 90 degrees and from 3 we have angle BPC is equal to angle A. Therefore angle 3 plus angle QPC is equal to 90 degrees. This is result 4. Now plus angle C is equal to angle A plus angle QPC since both have a pressure 90 degrees. The angle A cancels with angle A so this means angle QPC is equal to angle C. Now we try the QPC in this angle QPC is equal to angle C. Therefore QPC is equal to Q is equal to Q and VQ is equal to PQ Q is equal to QC. That is this BQ is equal to QC. This is what we were supposed to prove. If I should help you understand the solution of this question.