 Hello and welcome to the session. Let's work out the following question. It says D, E and F are respectively the midpoints of the sides AB, BC and CA of triangle ABC. Find the ratio of the areas of triangle D, E, F and triangle ABC. Let's now move on to the solution. We are given that D, E and F are the midpoints of the sides AB, BC and CA respectively. AB, BC and CA respectively. Therefore AB is twice of AB, BC is twice of BE or EC and similarly AC is twice of AF or FC. Now by a theorem we know that if a line is drawn through a midpoint of one side to meet the midpoint of the other side then this line is parallel to the third side and it is half of it. Therefore D, F is parallel to BC and it is half of BC. So D, F is one by two BC similarly DE is half of AC, EF is half of AB. Or we can say upon BC is equal to one by two DE upon AC is equal to one by two and EF upon AB is equal to one by two. Or we can say AB upon DE upon AC is equal to EF upon AB is equal to DF upon BC is equal to one by two. So the sides of the triangle ABC and DEF have the same ratio therefore triangle ABC is similar to triangle DEF. Now we know that the ratio of the areas of two similar triangles is equal to the square of the ratio there corresponding sides. Therefore area of triangle ABC upon area of triangle DEF is equal to two by one square since we have DE upon AC is equal to EF upon AB is equal to DF upon BC is equal to one by two. So ratio of area of triangle ABC and DEF would be two by one square but we have to find the ratio of the area of triangle DEF and ABC. Therefore it would be area of triangle DEF upon area of triangle ABC is equal to one by two square. So the ratio is one by four hence the answer is one is two four. This completes the question and the session. Bye for now. Take care. Have a good day.