If P be any point in the plane of square ABCD, prove that
`PA^(2)+PC^(2)=PB^(2)+PD^(2)`

If G be the centroid of a triangle ABC and P be any other point in the plane prove that PA^(2)+PB^(2)+PC^(2)=GA^(2)+GB^(2)+GC^(2)+3GP^(2)

Let ABCD is a square with sides of unit lenghts . Points E and F are taken on sides AB and AD respectively so that AE =AF . Let P be a point inside the square ABCD. The value of (PA)^(2)-(PB)^(2)+(PC)^(2)-(PD)^(2) is equal to -

P is a point inside the rectangle ABCD Prove that PA^2+ PC^2= PB^2 +PD^2 : .

P is an external point of the square ABCD. If PA gt PB ,then prove that PA^(2) - PB^(2) ~= PD^(2) - PC^(2)

Let ABCD be a rectangle and P be any point in its plane.Show that AP^(2)+PC^(2)=PB^(2)+PD^(2)

P is any point inside the rectangle ABCD. Prove that AP^2+CP^2=BP^2+DP^2

ABC is an equilateral triangle where AB = a and P is any point in its plane such that PA = PB + PC. Then (PA^(2)+PB^(2)+PC^(2))/(a^(2)) is

ABC is an equilateral triangle where AB = a and P is any point in its plane such that PA = PB + PC. Then (PA^(2)+PB^(2)+PC^(2))/(a^(2)) is

ABC is an equilateral triangle where AB = a and P is any point in its plane such that PA = PB + PC. Then (PA^(2)+PB^(2)+PC^(2))/(a^(2)) is

P is an interior point of square ABCD. Prove that PA+PB+PC+PD gt 2 AB .