 Hello and welcome to the session. In this session we discussed the following question which says, prove that the parallelogram, circumscribing the circle is the long bus. Before moving on to the solution, let's recall one fact which says that the length of the two tangents from an external point to circle equal key idea that we use for this question. Let's proceed with the solution now. We need to prove that the parallelogram circumscribing the circle is a long bus. So let's consider this figure in which we have centre O, B, C, D which is the parallelogram is a long bus. The sides of the parallelogram A, B, C, D touches the circle at the points P, Q, R and S that is the side A, B touches the circle at point P, side BC at Q, side CD at R and side AB at S. Now we have the circle with centre O, AS are the tangents drawn to the circle from point A. So we have AP is equal to AS. Let this be equation one. AP and AS are equal since these are the tangents to the circle from the external point A and so they are equal. Then we have BP is equal to BQ. Let this be equation two. They are equal as they are the tangents to the circle from the point B. Then we have CR is equal to CQ. Let this be equation three. AS these are the tangents from the external point C is equal to DS. Let this be equation four. They are equal as they are the tangents from the external point D to the circle. Now next adding equations one, two, three BP plus CR plus DR is equal to AS plus BQ plus CQ plus BP in one bracket plus CR plus DR in one bracket is equal to AS plus DS in one bracket plus BQ plus CQ in one bracket. Now from the figure you observe that AP plus BP is AB. So we have AB plus now CR plus DR is CD. So AB plus CD is equal to AS plus DS is AB plus BQ plus CQ is BC. Now this would be further equal to two times AB is equal to two times BC since we know that in a parallelogram opposite sides are equal therefore in the parallelogram they are equal therefore we have AB is equal to CD and BC is equal to AD. Hence AB plus CD can be written as two times AB and AD plus BC can be written as two times BC. This further means that we have AB is equal to BC. Now since AB is equal to CD BC is equal to AD therefore we get AB is equal to BC is equal to CD is equal to AD that is if we say CD is a rhombus that the parallelogram circumscribing the circle with center row is a rhombus that is ABCD is a rhombus. This completes the session hope you have understood the solution of this question.