 Hi and welcome to the session. Let us discuss the following question. Question says in figure 6.20, DE is parallel to OQ and DF is parallel to OR. Show that EF is parallel to QR. This is the figure 6.20. First of all let us understand basic proportionality theorem. Basic proportionality theorem states that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points the other two sides are divided in the same ratio. That is if we are given a triangle ABC in which DE is parallel to BC then by basic proportionality theorem we get ratio of these two sides is equal. So we can write AD upon DB is equal to AE upon EC. Now let us understand 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 in a triangle ABC DE is a line dividing AB and EC in the same ratio then DE is parallel to BC. This is converse of basic proportionality theorem. This is the key idea to solve the given question. Let us start with the solution. We are given this figure. Now in this figure let us consider triangle PQO. In triangle PQO we are given ED is parallel to QO or we can say DE is parallel to OQ. This is given in the question. This implies by basic proportionality theorem PE upon EQ is equal to PD upon DO. So we can write by basic proportionality theorem PE upon EQ is equal to PD upon DO. Let us name this expression as 1. Now let us consider triangle POR in triangle POR. DF is parallel to OR. This is given in the question itself. Now this implies by basic proportionality theorem PD upon DO is equal to PA upon FR. Let us name this expression as 2. Now clearly we can see in expression 1 and 2 PE upon EQ is equal to PD upon DO and PA upon FR is also equal to PD upon DO. So this implies PE upon EQ is equal to PA upon FR since both the ratios are equal to PD upon DO. So we can write from 1 and 2 we get PE upon EQ is equal to PA upon FR. Now let us consider triangle PQR. In triangle PQR clearly we can see PE upon EQ is equal to PA upon FR. This we have already shown above. Now by converse of basic proportionality theorem we get EF is parallel to QR. So our final answer is EF is parallel to QR. Hence proved this completes the session. Hope you understood the session. Take care and keep smiling.