 Hello and welcome to the session. I am Deepika here. Let's discuss a question which says how that the parallelogram circumscribing a circle is a robust. Now we know that the lengths of tangents drawn from an external point to a circle are equal. So this is a key idea behind our question. We will take the help of this key idea to solve the above question. So let's start the solution. Now we are given a parallelogram BCD whose sides touches a circle with center O. Now we have to prove ABCD is a rhombus. Now we know that if all sides of a parallelogram are equal then it is a rhombus. So we have to prove AB is equal to BC is equal to CD is equal to DA. Let PQRS be the points of content sides AB, BC, CD, DA respectively. Now lengths of two tangents drawn from an external point A to a circle are equal. Therefore AP is equal to AS. Let us give this as number one. Similarly we have lengths of two tangents drawn from an external point B to a circle are equal. That is BP is equal to BQ. We have CR is equal to CQ, DR is equal to DS. Let us give this as number two, this as number three and this as number four. Now on adding one, two, three and four we get AP plus BP plus CR plus DR is equal to AS plus BQ plus CQ plus DS. Now AP plus BP is AB plus DR is equal to CD is equal to, now AS plus DS is equal to AD, BQ plus CQ is BC. Now since AB, CD is a parallelogram therefore AB is equal to CD and BC is equal to AD. So we have twice AB equal to twice BC because AB, CD is a parallelogram. Therefore AB is equal to CD and BC is equal to AD. So this implies B is equal to BC. Now at distance size of a parallelogram are equal that is AB is equal to BC is equal to CD is equal to DA. AB, CD is A rhombus. Hence we have proved the parallelogram circumscribing a circle is A rhombus. I hope the solution is clear to you. Bye and take care.