prove OA*OA+OB*OB+OC* OC+ OD* OD =AB*AB+ BC*BC

Question

Parellogram ABCD,,Pt. E is pt. Of intersection of diagonals pt. O is in the plane of ABCD

Answer 1

We can apply the cosine rule to triangles AOD and AOB:

AD^2=AO^2+OD^2-2AO*ODcosAOD

AB^2=AO^2+OB^2-2AO*OBcosAOB.

But AD=BC, AO=OC, OD=OB, and AOD+AOB=180, so cosAOD=-cosAOB.

So, BC^2=OC^2+OD^2+2AO*OBcosAOB (from cosine rule for AOD).

Add the last equation and second equation together:

AB^2+BC^2=AO^2+OB^2+OC^2+OD^2

which is the same as OA^2+OB^2+OC^2+OD^2=AB^2+BC^2.