In a triangle A B C , the median from B to C A is perpendicular to the median from C to A B . If the median from A to B C is 30 , determine (B C^2+C A^2+A B^2 ) / 100 .

Step-by-step solution

Let G be centroid, D , E , F be mid-points of BC , CA , AB & M be mid-point of FE . Let BE =3 x , CF =3 y , given AD =30 Hence AM =5, GM =5, GD =10, BG =2 x , GE = x , CG =2 y , GF = y Now D is mid-point of hypotenuse of right angle triangle B G C . So D is circum centre of the triangle. SO BD = GD =10 BC =20 Hence 4 (x^2+y^2 )=400 x^2+y^2=100 Now 9 (x^2+y^2 )= 1 4 2 B C^2+2 A B^2-A C^2+2 B C^2+2 A C^2-A B^2 aligned & 900 4=4 B C^2+A B^2+A C^2 & A B^2+A C^2=3600-1600=2000 aligned Hence A B^2+B C^2+C A^2 100 = 2400