 Hello and welcome to the session. Let's discuss the following question. It says if the areas of two similar triangles are equal, prove that they are congruent. So let's now move on to the solution and let's first write what is given to us. We have given two similar triangles having same areas. So we take two triangles as ABC that is similar to triangle say DEF such that they have same area that is area of triangle ABC is equal to area of triangle DEF and what we have to prove? We have to prove that these triangles are congruent that is triangle ABC is congruent to triangle DEF. So let's now start on with the proof. Now we are given that triangle ABC is similar to triangle DEF. Now we know that if two triangles are similar then their corresponding sides are in the same ratio and the corresponding angles are equal. So we have angle A is equal to angle D, angle B is equal to angle E, angle C is equal to angle F and their corresponding sides are in the same ratio. So we have AB upon DE is equal to BC upon EF is equal to CA upon FD. Now we also know that the ratio of the areas of two similar triangles is equal to the square of ratio of their corresponding sides. So also since triangle ABC is similar to triangle DEF therefore ratio of the areas that is area of triangle ABC upon area of triangle DEF is equal to the square of ratio of their corresponding sides that is AB square upon DE square is equal to BC square upon EF square is equal to CA square upon FD square and this is by the result the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Let us name this as one. Now we know that area of triangle ABC is equal to area of triangle DEF. So we have one is equal to AB square upon DE square is equal to BC square upon EF square is equal to CA square upon FD square as area of triangle ABC is equal to area of triangle DEF. So it gets cancelled and we have one is equal to AB square upon DE square is equal to BC square upon EF square is equal to CA square upon FD square. Now considering these two we have AB square is equal to DE square and taking square root we have AB is equal to DE similarly BC square upon EF square is equal to one implies BC square is equal to EF square I am taking square root we have BC is equal to EF and similarly we have CA square upon FD square is equal to one implies CA square is equal to FD square and this implies CA is equal to FD. Hence triangle ABC is congruent to triangle DEF as all the three sides of one triangle are equal to the corresponding sides of the other triangle. So this is by SSS congruence condition which says that two triangles are congruent if the three sides of one triangle are equal to the three sides of the other triangle hence triangle ABC is congruent to triangle DEF hence the result is proved. So this completes the question and the session. Bye for now take care have a good day.