P is a point inside the rectangle ABCD Prove that `PA^2+ PC^2= PB^2 +PD^2`:
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Suppose a^2,b^2,c^2 are in A.P. Prove that 2b^2=a^2+c^2

Suppose that a, b, c are in A.P. Prove that 2b = a + c

If A and B are the points (3,4,5) and (-1,3,-7) respectively, find the equation of the set of points P such that PA^2+PB^2=k^2 where k is a constant.

Consider the points A(-5,6,8), B(1,8,11),C(4,2,9) and D(-2,0,6). Find the mid points of AC and the mid point of BD and prove that ABCD is a parallelogram

Let x and y be the length and breadth of the rectangle ABCD in a circle having radius r. Let angleCAB=theta (Ref. Figure). If triangle represent area of the rectangle and r is a constant. Find (d triangle)/(d theta) and (d^2 triangle)/(d theta^2) .

In the parallelogram ABCD , the line drawn through Q point P on AB , parallel to BC meets AC at Q . The line through Q , parallel to AB meets AD R . Prove that (AP)/(PB)=(AR)/(RD) . .

The vertices of the rectangle ABCD are A(-1, 0), B(2, 0), C(a, b) and D(-1, 4) Then, the length of the diagonal AC is

Suppose a, b, c are in A.P. Prove that bc-a^2+ab-c^2=2b^2-(a^2+c^2) . Hence, Prove that bc-a^2,ca-b^2,ab-c^2 are in A.P.