 Hi and welcome to the session. I am Shashi and I am going to help you with the following question. Question says, using theorem 6.2 prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side. First of all let us understand theorem 6.2. Theorem 6.2 is converse of basic proportionality theorem. Converse of basic proportionality theorem states that if a line divides any two sides of a triangle in the same ratio then the line is parallel to the third side. That is, if we are given a triangle abc de divides ab and ac in the same ratio then de is parallel to bc. If ad upon db is equal to ae upon ac then de is parallel to bc. This is the key idea to solve the given question. Now let us start the solution. Let us consider a triangle abc in which d and e are midpoints of ab and ac. We have to prove that de is parallel to bc. Now let us write given a triangle abc de and e are midpoints of ab and ac. Now we have to prove that de is parallel to bc. Now let us start the required proof. Now we know de is the midpoint of ab. So this implies ad is equal to db. If we find out ratio ad upon db then it must be equal to 1 since ad is equal to db. Let us name this expression as 1. Now we know e is the midpoint of ac. It is given in the question. So ae is equal to ac. Now if we find out the ratio ae upon ac then it must be equal to 1 as ae is equal to ac. Now let us name this expression as 2. Now from expression 1 and expression 2 we get ad upon db is equal to ae upon ac is equal to 1. ad upon db is equal to ae upon ac implies de is a line which divides ab and ac in the same ratio. So by converse of basic proportionality theorem we get de is parallel to bc. Hence we have proved that line joining the midpoints of two sides of a triangle is parallel to the third side. So this is our required proof. This completes the session. Hope you understood the session. Take care and have a nice day.