 Hello and welcome to the session. Let us understand the following question today. In an equilateral triangle A, B, C, D is a point on side B, C such that B, D is equal to 1 by 3 of B, C. Prove that 9 A, D square is equal to 7 A, D square. Now let us write the solution. It is given to us that an equilateral triangle A, B, C and D is a point on B, C such that B, D is equal to 1 by 3 of B, C. We have to prove 9 A, D square is equal to 7 A, B square. Let us see the figure. Here we have equilateral triangle A, B, C, D is some point on B, C such that B, D is equal to 1 by 3 of B, C. And we have to prove 9 A, D square is equal to 7 A, V square. Here we will be having a construction. Draw A, E perpendicular to B, C which meets B, C at E. Draw A, E perpendicular to B, C which meets B, C at E. Now let us solve it further. Let us write the proof. Since by construction A, E is perpendicular to B, C which implies B, E is equal to half of B, C which is equal to half of A, B. Let us name it as number 1. This is because A, B is equal to B, C is equal to A, C in equilateral triangle A, B, C. So now in a right-angle triangle A, B, E by Pythagoras theorem we have A, B that is the hypotenuse A, B square is equal to B square plus A, E square. That is A, B square is equal to B is nothing but half of A, B so replacing it by A, B by 2 the whole square plus A, E square as it is. A square is equal to A, B square minus 1 by 4 A, B square or A square is equal to 3 by 4 A, B square. Let us name it as number 2. Now in right-angle triangle A, D, E by Pythagoras theorem we have A, D square is equal to A square plus D square or we can write A square is equal to A, D square minus D square. And let us name it as number 3. Now it is given to us that D trisects B, C that is B, D is equal to 1 by 3 B, C. B, D is equal to 1 by 3 A, B. Let us name it as number 4. This is because A, B is equal to B, C in an equated triangle A, B, C. So now therefore B, E minus B, D is equal to we have B, E is equal to 1 by 2 A, B by 1. 1 by 2 A, B minus B, D is equal to 1 by 3 A, B by 4. Therefore B, E minus B, D. We can see the figure B, E minus B, D is equal to D, E. So therefore B, E is equal to 1 by 6 A, B. Let us name it as number 5. Now 5 in equation 3 we get we have our equation 3 as A, E square is equal to A, D square minus D square. So we write A, E square is equal to A, D square minus D square. That is A, E square is equal to A, D square minus instead of D we will substitute the value of D as A, B by 6 the whole square. This is our equation 6. Now from 2 and 6 we get we have our equation 2 as A, E square is equal to 3 A, B square 3 by 4 A, B square. So substitute A, E square as 3 by 4 A, B square. 3 by 4 A, B square is equal to A, D square minus 1 by 6 A, B square. That is 3 by 4 A, B square is equal to A, D square minus 1 by 36 A, B square. Or 3 by 4 A, B square plus 1 by 36 A, B square is equal to A, D square. 27 A, B square plus A, B square is equal to 36 A, D square. Or 28 A, B square is equal to 36 A, D square. Dividing both sides by 4 we get 7 A, B square is equal to 9 A, D square. And this is our required result. Hence proved, I hope you understood the question. Bye and have a nice day.