 Hi, and welcome to another session where we will discuss the following question. The question says, A, B, C is a triangle, O is the circum-center of triangle A, B, C versus P, Q, R acting along O, A, O, V, O, C are in equilibrium. Prove that P is to Q is to R is equal to A square into B square plus C square minus A square is to V square into C square mu plus A square minus V square is to C square into A square plus V square minus C square. Now, we begin with the solution. We are given that forces P, Q, R along the lines O, A, O, B, O, C is the circum-center of triangle A, B, C. Since O is the circum-center of triangle A, B, C, therefore, angle B, O, C is equal to 2A, angle C, O, A is equal to 2B, and angle A, O, B is equal to 2C. Now, since forces are in equilibrium, therefore, by Lamy's theorem, E by sin of angle B, O, C is equal to Q by sin of angle C, O, A is equal to R by sin of angle A, O, B. Now, C is equal to 2A, so this implies P by sin 2A is equal to Q by sin 2B is equal to R by sin 2C. This implies P by 2 sin A cos A is equal to Q by 2 sin B cos B is equal to R by 2 sin C cos C. Now, let us assume that sin A by A is equal to sin B by B is equal to sin C by C and let it be equal to C. This implies B by V square plus C square minus A square by 2VC is equal to Q by KV into C square plus A square minus B square divided by 2AC is equal to R by AC into A square plus B square minus C square divided by 2AB. This implies P by A square into B square plus C square minus A square is equal to Q by B square into C square plus A square minus B square is equal to R by C square into A square plus B square minus C square. And this implies P is to Q is to R is equal to A square into B square plus C square minus A square is to B square into C square plus A square minus B square is to C square into A square plus B square minus C square. So, this completes the session. Bye and take care.