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

Let ABCD be a rectangle and P be any point in its plane. Show that PA^2+PC^2=PB^2+PD^2 using coordinate geometry.

Let ABCD be a square of side length 2 units. C2 is the circle through vertices A, B, C, D and C1 is the circle touching all the sides of the square ABCD. L is a line through A. . If P is a point on C1 and Q in another point on C2, then (PA^2+ PB^2+ PC^2+PD^2)/ (QA^2+ QB^2+ QC^2+ QD^2) is equal to

If A and B be square matrices of the same order such that AB=BA, prove that : (A-B)^2 = A^2 - 2AB + B^2

O is any point inside a rectangle ABCD (see Fig. 6.52). Prove that OB^2 + OD^2 = OA^2 + OC^2 .

Of all the line segments drawn from a point P to a line l not containing P, let PD be the shortest. If B and C are points on l such that D is the mid-point of BC, prove that PB = PC.

If z is any complex number, prove that : |z|^2= |z^2| .

If p^(2) = a^(2) cos^(2) theta + b^(2) sin^(2)theta , prove that p + (d^(2p)/(d theta^(2))) =(a^(2)b^(2))/p^(3)

In any triangle ABC, prove that : (tanC)/(tanA)= (b^2+c^2-a^2)/(a^2+b^2-c^2) .

Let P(a, b, c) be any on the plane 3x+2y+z=7 , then find the least value of 2(a^2+b^2+c^2) .

If the diameter through any point P of a parabola meets any chord in A and the tangent at the end of the chord meets the diameter in B and C, then prove that PA^2=PB.PC