The OABC is a tetrahedron such that `OA^2+BC^2=OB^2+CA^2=OC^2+AB^2`,then

If G be the centroid of a triangle ABC, prove that, AB^2 + BC^2 + CA^2 = 3(GA^2 + GB^2 + GC^2)

In a tetrahedron OABC, the edges are of lengths, |OA|=|BC|=a,|OB|=|AC|=b,|OC|=|AB|=c. Let G_1 and G_2 be the centroids of the triangle ABC and AOC such that OG_1 _|_ BG_2, then the value of (a^2+c^2)/b^2 is

In an equilateral triangle ABC, BE is perpendicular to side CA. Prove that: AB^(2) + BC^(2) +CA^(2) = 4BE^(2)

OABC is a tetrahedron where O is the origin and A,B,C have position vectors veca,vecb,vecc respectively prove that circumcentre of tetrahedron OABC is (a^2(vecbxxvecc)+b^2(veccxxveca)+c^2(vecaxxvecb))/(2[veca vecb vecc])

In the triangle OAB , M is the midpoint of AB, C is a point on OM, such that 2 OC = CM . X is a point on the side OB such that OX = 2XB. The line XC is produced to meet OA in Y. Then (OY)/(YA)=

If AD, BE and CF are the medians of a Delta ABC, then evaluate (AD^(2)+BE^(2)+CF^(2)): (BC^(2) +CA^(2) +AB^(2)).

If OAB is a tetrahedron with edges and hatp, hatq, hatr are unit vectors along bisectors of vec(OA), vec(OB):vec(OB), vec(OC):vec(OC), vec(OA) respectively and hata=(vec(OA))/(|vec(OA)|), vecb=(vec(OB))/(|vec(OB)|), vec c= (vec(OC))/(|vec(OC)|) , then :

Diagonals of rhombus ABCD intersect each other at point O. Prove that : " "OA^(2)+OC^(2)=2AD^(2)-(BD^(2))/2

The point O is outside the angleABC . Show that 2(OA+OB+OC) gt AB+BC+CA .

A(1, -2) and B(2, 5) are two points. The lines OA, OB are produced to C and D respectively, such that OC=2OA and OD=2OB . Find CD .