 Hello and welcome to the session. In this session we discussed the following question which says two tangents P a and P b are drawn to a circle with centre O from an external point P, proves that angle a P b is equal to two times angle O a b. Before we move on to the solution let's recall some facts. First we have the tangent to a circle is perpendicular to the radius through the point of contact. Then the other result is the length of two tangents from an external point to a circle are equal. Then we also have one more result. In a triangle the angles opposite equal sides are equal. This is the key idea that we use in this question. Let's proceed with the solution now. Consider this figure. Here we are given the two tangents P a and P b are drawn from the point P to the circle with centre O. That is we are given that P a and P b two tangents to the circle the external point or from the point P and we need to prove that angle a P b that is this angle is equal to two times angle O a b that is this angle. Let's start with the proof now. First to call let's mark these angles. Let this be angle one, this be angle two, this be angle three and this be angle four. Now since we have P a and P b are the tangents so P a would be equal to P b. Since these are the tangents from the external point P and so they are equal. As we know that the lengths of the tangents from an external point to a circle are equal. So in the triangle P a b as P a is equal to P b so this means that angle three would be equal to angle four as they are the angles opposite the equal sides and we know that in a triangle the angles opposite the equal sides are equal so angle three is equal to angle four as P a is equal to P b. Now as we know that the tangent to a circle is perpendicular to the radius through the point of contact so O a is perpendicular to the tangent P a therefore we have angle O a P is equal to 90 degrees. Tangent is perpendicular to the radius through the point of contact. Now since angle O a P is equal to 90 degrees so this means angle one plus angle three is equal to 90 degrees. So from here we have angle three is equal to 90 degrees minus angle one consider the triangle P a b in this we have angle two plus angle three plus angle four is equal to 180 degrees by the angle some property of a triangle that is the sum of the three angles of a triangle is equal to 180 degrees. Now we have angle three is equal to angle four so here we have angle two plus two times angle three is equal to 180 degrees since we have angle three equal to angle four. Now substituting angle three equal to 90 degrees minus angle one in this equation we get angle two plus two times 90 degrees minus angle one is equal to 180 degrees since we have angle three equal to 90 degrees minus angle one. So further we get angle two plus 180 degrees minus two times angle one is equal to 180 degrees now 180 degrees and 180 degrees cancels so we get this is equal to two times angle one equal to angle two. Now from the figure you can see that angle one is equal to angle O a b so we have two times angle O a b is equal to angle two which is angle a p b so hence proved angle O a b is equal to angle a p b or you can say that angle a p b is equal to two times angle O a b. This is what we were supposed to prove so this completes the session hope you have understood the solution of this question.