Proof with circles?

Question

P is any point on the diameter AB of a circle. CD is any chord of the circle parallel to AB. Prove that (PC^2)+(PD^2) is independent of the position of CD.

My friend and I tried several ways to do this question through cartesian system and analytical method and even with vectors but we couldnt come up with a solid proof.

Advice and help greatly appreciated.

Answer 1

This proof is surprisingly simple using analytic geometry. Place a circle of radius r at the origin and place the diameter AB along the x-axis. The chord, parallel to AB has a constant dimension yc. The endpoints of the chord CD are:

C = (√(r^2 - yc^2), yc) and D = (-√(r^2 - yc^2), yc)

The x coordinates are determined by x^2 + y^2 = r^2

Now choose any point P along the diameter P = (xp,0)

The square of the distances from P to C and D are the sum of the squares of the coordinate differences:

PC^2 = (Px - Cx)^2 + (Py - Cy)^2 = (xp - √(r^2 - yc^2))^2 + yc^2
PC^2 = xp^2 - 2xp √(r^2 - yc^2) + (r^2 - yc^2) + yc^2
PC^2 = xp^2 - 2xp √(r^2 - yc^2) + r^2

Similarly:

PD^2 = (Px - Dx)^2 + (Py - Dy)^2 = (xp + √(r^2 - yc^2))^2 + yc^2
PD^2 = xp^2 + 2xp √(r^2 - yc^2) + (r^2 - yc^2) + yc^2
PD^2 = xp^2 + 2xp √(r^2 - yc^2) + r^2

Add the two together:

PC^2 + PD^2 = (xp^2 - 2xp √(r^2 - yc^2) + r^2) + (xp^2 + 2xp √(r^2 - yc^2) + r^2
PC^2 + PD^2 = 2(xp^2 + r^2)

The result is independent of yc which is the position of the chord. So it is independent of the posiiton of the chord.

Answer 2

Let the center of the circle be A. Let there be a line extending from the top of the circumference to the bottom, the top point being B and the bottom being E. Let there be a chord line CD, where C and D are random points taken on the circumference of the circle but such that the straight line joining CD is perpendicular to the straight line extending from B to E and runs through center A. Therefore the cutting angles between chord AB, and CD is 90 degrees, because a straight line set on a straight line makes a 90 degrees angle. Further lets say the point of intersection between chord AB and CD is F. Because similar angles subtend similar sides of equal length, CF=DF and AF=AF because these are all subtended by the similar angles 90 degrees. If in this triangle within the circle there are 2 sides of equal length and and angle of equal proportion subtending, it means the third length must also be equal as well as there angles are the same. Therefore AC=AD. As a result AC^+AD^2. Now if the point is moved on the chord BE, CF will still equal DF and the side lengths are still going to be subtended by 90 degrees because straight line upon straight line still. Now A can replaced with any point on diameter CD and AC^2+AD^2. This is true because CHORD CD IS PERPENDICULAR TO BE BUT YOUR PROOF OF CD BEING PARREL TO EB DOES NOT HOLD FOR THIS PROOF.

Answer 3

Scale the circle to radius=1, put the diameter AB on the x axis and point P=(x,0). Put chord CD as the line y=k>0 so points C and D are (+/- sqrt(1-k^2) ,k). Then PC= sqrt( (x-sqrt)^2 +k^2 ) and PD= sqrt( (x+sqrt)^2 +k^2 ) so then PC^2+PD^2 = (x-sqrt)^2 +k^2 +(x-sqrt)^2 +k^2 then expand to get 2x^2+2 which is independent of k

Answer 4

You might try proving it by contradiction. Assume that (PC^2) + (PD^2) is dependent on the position of CD. All you need is one counterexample to prove the statement wrong and you've proven independence by contradiction. Remember, that a diameter is also a chord and that a line (or segment thereof) can be parallel to itself. So, CPD could be a single line or a triangle. If I come up with anything better, I'll edit this answer.

Answer 5

I'm afraid that proof is invalid. (PC^2) + (PD^2) is not a constant but will change with the position of P. The law of cosines says:

(CD^2) = (PC^2) + (PD^2) - 2(PC)(PD)cos P

Therefore

(PC^2) + (PD^2) = (CD^2) - 2(PC)(PD)cos P

The length CD is a constant. If (PC^2) + (PD^2) is also a constant as has been proposed, then 2(PC)(PD)cos P must be a constant as well.

If angle P is a right angle then cos P = 0 and
2(PC)(PD)cos P = 0.
If it is not a right angle, then 2(PC)(PD)cos P ≠ 0. And it will not always be a right angle.

So the proof is invalid.

Answer 6

A-X