 Hello and welcome to the session. I am Asha and I am going to help you with the following question that says if EFGH are respectively the midpoints of sites of a parallelogram ABCD show that area of EFGH is equal to half area of ABCD. So first let us learn a simple result which says if a triangle and a parallelogram are on the same base between the same parallel lines then EFGH equal to half the area of parallelogram. Use this result. Suppose this is a parallelogram we have a triangle on the same base and between the same parallel lines. Suppose ABCD is the parallelogram and on the base DC a triangle stands which touches the line AB at the point E. So the triangle and the parallelogram are on the same base and between the same parallel lines AB and CD. Now let us prove that area of triangle is half the area of parallelogram. Now area of triangle as we know is half of base into height and the other triangle is DEC which is equal to half base is DC and let the perpendicular distance between the two parallel lines be denoted by H. So this is the area of triangle half of base into height. Now this is equal to half of area of parallelogram ABCD since area of a parallelogram is the product of base into the corresponding altitude. So this is the required result which shows that area of triangle is equal to half area of parallelogram. So with the help of this result we are going to prove the above question. So this is our key idea. Let us now interpret the given question in the form of a figure and the figure will look like this where E, F, G and H are respectively the midpoints of AB, BC, CD and DI which are the sides of the parallelogram. So we are given a parallelogram ABCD in which E, F, G and H are midpoints of AB, BC, CD and DI respectively and we have to show that area of EFGH is equal to half area of ABCD. Now in quadrilateral HD, CF, HD is equal to half of AD since H is midpoint of AD. So this implies AH is equal to HD is equal to half of AD. This is further equal to half of BC since opposite sides of the parallelogram are equal. So AD is equal to BC and half of BC is further equal to FC since again F is midpoint of side BC. So this implies that HD is equal to FC. Also HD is parallel to FC since AD is parallel to BC that is opposite sides of a parallelogram are equal so this implies a part of this line is also parallel to a part of this line. Therefore we have HD parallel to FC. This implies that quadrilateral HD CF is a parallelogram since in a quadrilateral if a pair of opposite sides is equal and parallel then it is a parallelogram therefore in this quadrilateral we have a pair HD and HD which is equal also and parallel also therefore it is a parallelogram. Now in a parallelogram opposite sides are equal and parallel so this implies HF is parallel to DC. So let this be equation number one. Now from the figure as we can see the triangle HGF and the parallelogram HD CF are on the same base HF and between the same parallel lines HF and DC so this implies area of triangle HDF is half area of parallelogram HD CF by our key idea that is area of triangle HGF is equal to half area of parallelogram HD CF. And let this be equation number two. Now AB is parallel to DC since opposite sides of a parallelogram are parallel and equal also from one we have HF parallel to DC so this implies two lines which are parallel to the same even line are parallel to each other so this implies HF is parallel to AB. Now applying the same key idea on the triangle EHF and the parallelogram AHFB which are on the same base HF and between the same parallel lines AB and HF we can write area of triangle HEF is equal to half into area of parallelogram ABFH and let this be equation number three. From equation two and three on adding them we have on the left hand side area of triangle HDF plus triangle HEF and on the right hand side we have half area of parallelogram HDCF plus half area of parallelogram ABFH. Now area of HDF plus HEF which are these two triangles is nothing but area of parallelogram EHF fine so this is area of parallelogram EFGH and on the right hand side we have half any common an area of parallelogram HDCF which is this parallelogram and area of parallelogram ABF which is the above one and the sum of both these parallelograms is the whole of ABCD which is the given parallelogram so this is equal to half area of ABCT and thus we have area of EFGH equal to half area of ABCT. So this completes the session take care and have a good day.