 Hello and welcome to the session. I am Asher and I am going to help you with the following question which says, ABC is a triangle right angled at C, aligned through the midpoint, M of the hypotenuse AB and parallel to BC intersects AC at D. Show that first D is the midpoint of AC, second MD is perpendicular on AC, third CM is equal to MA is equal to half of AB. Let us now begin with the solution and we are given triangle ABC in which angle C is equal to 90 degree, M is midpoint of AB and a line through M is drawn parallel to BC such that it intersects AC at D. So this implies MD is parallel to BC and first we have to show that D is the midpoint of AC. Now in triangle ABC, M is midpoint of AB, MD is parallel to BC. So this implies that D is midpoint of AC and this is the theorem 8.10 of your book which says the line drawn through the midpoint of one side of a triangle parallel to another side by six third side. And here the line is drawn from the midpoint of AB that is from the midpoint M parallel to side BC so it will bisect the third side which is side AC and this implies that D is the midpoint of AC. So this completes the first part and next we have to show MD is perpendicular on AC. Now angle C is 90 degree in triangle ABC, MD is parallel to BC and let AC be the transversal and this implies that angle C is equal to angle ADM since when two parallel lines are intersected by a transversal then the corresponding angles are equal. So these two are corresponding angles and angle C is given to us as 90 degree. So this implies angle ADM is equal to 90 degree that is this angle is also 90 degree and since this is a straight line so this is also 90 degree and this further implies that MD is perpendicular on AC. So this completes the second part also and now we have to show that is equal to ME is equal to half of AB. So first let us try CM now in triangle ADM and triangle CDM. In the first part we have proved that D is the midpoint of AC so this implies AD is equal to DC this is from one angle DM is equal to angle CDM is equal to 90 degree each and this is from part two since we have proved that MD is perpendicular on AC so this implies both these angles are of 90 degree each and MD is equal to DM since the side is common to both the triangles therefore by side angle side congruence relation triangle ADM is congruent to triangle CDM. So this implies AM is equal to CM this is by CPCD that is corresponding to two parts of congruent triangles are equal so let this be equation number one and now since M is the midpoint of AB so this implies AM is equal to MD is equal to half of AB let this be equation number two. Now from one and two we have CM is equal to AM or MA and AM is equal to half of AB and this is what we have to prove thus we can say that CM is equal to MA is equal to half of AB so this completes the third part and hence the session hope you enjoyed it take care and have a good day.