 Hello and welcome to the session. In this session we discussed the following question which says, in the given figure, pq is parallel to rs and a transversal ab cuts them at c and d respectively. If cl and dm are the bisectors of the alternate angles, angle pcd and angle cds respectively means that cl is parallel to dm. We know that if two parallel lines are intersected by a transversal, then each pair of alternate interior angles are equal. This is the key idea to be used for this question. Now we move on to the solution. First let's see what all is given to us. In this we have pq is parallel to rs and this ab is the transversal, then cl is the bisector of angle pcd, dm is the bisector of angle cds. We also have angle pcd and angle cds are alternate angles. Then we need to prove cl is parallel to dm. Now that we have since pq is parallel to rs, ab is a transversal, then this means that angle pcd is equal to angle cds as they are the alternate angles or alternate interior angles, so they are equal. As we know that if two parallel lines are intersected by a transversal, then each pair of alternate interior angles are equal. So this would mean that half of angle pcd is equal to half of angle cds. Now since we know that cl is the bisector of angle pcd, so we have angle pcl is equal to angle lcd since cn is the bisector of angle pcd. Then we have angle cdm is equal to angle mds since we know that dm is the bisector of angle cds. So therefore half of angle pcd equal to half of angle cds means that angle lcd is equal to angle cdm. That is these two angles are equal. Now from the figure you can see that the lines lc and dm are intersected by the transversal cd, angle lcd is equal to angle cdm. That is the pair of alternate interior angles are equal. Therefore this would mean that cl is parallel to dm. The reason for this is that if two lines are intersected by a transversal such that the angles of any pair of alternate angles or alternate interior angles are equal then the two lines are parallel. So hence we have proved that cl is parallel to dm. So this completes the session. Hope you have understood the solution for this question.