 Hi and welcome to the session. Let us discuss the following question. Question says, using theorem 6.1 prove that a line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side. First of all let us understand theorem 6.1. Theorem 6.1 is also known as basic proportionality theorem. It states that if a line is drawn parallel to one side of a triangle intersecting other two sides in distinct points then the other two sides are divided in the same ratio. That is if we are given a triangle PQR such that ST is parallel to QR then PS upon SQ is equal to PT upon TR. We know ST is parallel to QR ST intersects PQ and PR in distinct points so it will divide PQ and PR in the same ratio. This is the key idea to solve the given question. Now let us start with the solution. First of all let us consider a triangle ABC and mark the midpoint of AB as D. Through D draw a line parallel to BC intersecting AC at E. So we can write that we are given a triangle ABC in which D is the midpoint of AB and DE is parallel to BC. Now we have to prove that DE bisects AC or we can say we have to prove that AE is equal to AC. Let us start with the proof now. Now clearly we can see in triangle ABC DE is parallel to BC this is given. Now by basic proportionality theorem we get AD upon DB is equal to AE upon AC. We know DE is parallel to BC and also DE is intersecting AB and AC at distinct points D and E. So DE will divide AB and AC in the same ratio. So we can write by basic proportionality theorem AD upon DB is equal to AE upon AC. Let us name this expression as 1. Now we also know that AD is equal to DB since D is the midpoint of AB. Now let us name this expression as 2. Now substituting this value of AD in expression 1 we get DB upon DB is equal to AE upon AC. Now DB and DB will cancel each other and we will get 1 on left hand side. So we can write 1 is equal to AE upon AC. Now multiplying both the sides by EC we get EC is equal to AE. Now EC is equal to AE implies E is the midpoint of AC. So we can say DE bisects AC. We were required to prove this only. So we can say DE bisects AC hence proved this completes the session. Hope you understood the session. Take care and have a nice day.