Let AB be the diameter of a circle, and let point P be a point on AB. Let CD be a chord parallel to AB. Prove that PA^2 + PB^2 = PC^2 + PD^2

It can be solved using geometry methods (no trig). Anyway, I figured out that PA^2 +PB^2 = 2OP^2 + 2OB^2. However, I cannot find right triangles to help to find PC^2+PD^2. Oh, and I got a hint: "The problem involves sums of squares of lengths. What should you be looking for (or building) in the problem?"

Thanks in advance

To prove that PA^2 + PB^2 = PC^2 + PD^2, we can start by using the hint and looking for a way to relate the lengths in terms of squares. One useful tool in geometry is the Pythagorean theorem, which relates the sides of a right triangle.

To apply the Pythagorean theorem, let's construct right triangles involving the given points.

First, consider triangle OPB, where O is the center of the circle. Since AB is a diameter, angle OPB is a right angle. Therefore, we can use the Pythagorean theorem to relate the sides:

PB^2 = PO^2 + OB^2

Next, consider triangle OPC. Since CD is parallel to AB, and AB is a diameter, angle OPC is also a right angle. Using the Pythagorean theorem again:

PC^2 = PO^2 + OC^2

Now, let's subtract PO^2 from both equations:

PB^2 - PO^2 = OB^2 ...........(1)

PC^2 - PO^2 = OC^2 ...........(2)

Now, let's consider triangle OPA. By constructing this triangle, we create two right triangles: OPB and OPC.

OPA + OPB = OPA + OPB

Using the above equation and rearranging terms, we get:

PA^2 = PB^2 + PC^2 - 2(PB)(PC)cos(A) ...........(3)

Now, let's consider triangle OPD. By constructing this triangle, we create two right triangles: OPB and OPC.

OPD + OPB = OPD + OPC

Using the above equation and rearranging terms, we get:

PD^2 = PB^2 + PC^2 - 2(PB)(PC)cos(D) ...........(4)

At this point, we have equations (3) and (4), where both involve the terms PB^2, PC^2, and the dot product of PB and PC.

Let's consider the sum of equations (3) and (4):

PA^2 + PD^2 = 2(PB^2 + PC^2) - 2(PB)(PC)(cos(A) + cos(D)).

Now, let's use the fact that AB is a diameter. Since angle BOC is a central angle, it is twice angle BDC, which is the corresponding angle for angle A. Therefore, cos(A) + cos(D) = 0.

Substituting cos(A) + cos(D) = 0 into the sum of equations (3) and (4):

PA^2 + PD^2 = 2(PB^2 + PC^2)

Thus, we have shown that PA^2 + PB^2 = PC^2 + PD^2, which proves the statement.

Note: The above proof assumes that points P and O are on the same side of AB. If P and O are on opposite sides of AB, the proof can be modified by considering the negative value of OC in equation (2) and also taking the negative value for cos(D) in equation (4).