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Power of a Point Proof

  1. Jul 15, 2007 #1
    I am trying to rigorously prove the POP theorem. Here is perhaps the best one available: http://www.cut-the-knot.org/pythagoras/PPower.shtml

    Can someone link me to a more complete proof?

    A good proof would need to cover all of the cases: when P is inside the circle, on the circle (a rather obvious case), or outside the circle, and when the lines are tangents, secants, or chords.

    Also, I think that link uses a geometric theorem that goes something like angles intercepting the same arc are equal, which I am also having difficulty proving. So, if someone could link me to a proof of that or provide a short one here that would be great.
  2. jcsd
  3. Jul 15, 2007 #2

    Gib Z

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    I've never heard of this "Power of a point" theorem, although may because my textbook for some very stupid reason has not named its theorems in the Circle geometry section, or because I haven't finished the chapter yet, but this will be good practice for my Circle Geometry test, so yay :)

    Ok well I will prove them in the cases shown on your link.
    For the case of the first (far left) picture:

    2 chords AD and BC intersect at P. ( Data)

    In Triangles APC and DPB, angle CAP = angle BDP since angles at the circumference standing on the same arc are equal.

    Angle APC = Angle DPB (Vertically opposite)

    Hence Triangles APC and DPB are similar triangles (equiangular)

    Hence, AP/PD = CP/PB (matching sides of similar triangles are in ratio).

    [itex]AP \cdot PB = CP \cdot PD[/itex]

    That is, the products of the intercepts of two intersecting chords of a circle are equal.

    For the Case of the Second (middle) picture:

    The secants PA and PC cut the circle at B and D respectively. (Data)

    In triangles ADP and CBP, angle BAD= angle BCD (angles at the circumference standing on the same arc are equal).

    angle APD=angle CPB (common angle).

    Triangles ADP and CBP are similar (equiangular).

    Hence, PA/PC = PD/PB, or [itex]PA\cdot PB = PC\cdot PD[/itex]
    For the last case:

    PT is a tangent, PA is a secant, which cuts the circle at B. (Data)

    In triangles BPT and TPA,
    angle PTB = angle TAB (angle in the alternate segment)
    angle TPB = angle TPA (common angle)
    Hence triangles BPT and TPA are similar (equiangular).

    PT/PA = PB/PT (matching sides of similar triangles are in ratio).

    [itex]PT^2 = PA \cdot PB[/itex]
  4. Jul 17, 2007 #3

    Gib Z

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    Ahh sorry as you may have noticed, that proof doesn't seem to work out because I labeled some of the points wrong. The proof works out fine if you copy down the pictures off the link, and for the first one: Replace all B's with C's, and C's with B's, ie swap B and C.
    For the second one: Replace D's with A's, A with B, C with D and B with C.
    Third one: B with A, and C with B, A(d) with T.

    Hope it wasn't too much of an inconvenience.
  5. Jul 17, 2007 #4
    Your POP question ringed a bell, though not at first sight. I looked it up in a book on plane geometry as used for the study for college teachers in the 1920's. I don't think it is actual these days and that means it is hard to find. You succeeded and the prove you found is the correct one. There really isn't much more to say about it. Only missing might be: AP*PD=CP*PB is the same as AP:CP=PB:PD. All you have to do is to look for the similar triangles [itex]\triangle{} APB[/itex] and [itex]\triangle{} CPD[/itex]. Out of their similarity follows AP:CP=PB:PD. For P inside the circle the power of P is negative, for P on the circle it is 0 and outside the circle positive.
    The consequences of this theorem are more known than the theorem itself, by the way.
  6. Jul 17, 2007 #5
    In addition to the above given remarks: If you choose P, say, inside the circle than any two lines that are intersecting in P constitute the same equality. So it is meaningful to speak of THE power of P. More, you can draw a concentric circle through M and P and all points on that circle have the same power as P does.The theorem is mostly used to construct a piece of line when three pieces of line are given and x has to be equal to [itex]\frac{ab}{c}[/itex].
  7. Jul 17, 2007 #6
    This is the part that is not exactly obvious to me. It is probably a well-established theorem but can someone anyway link me to a proof or provide a quick one here. Thanks.
  8. Jul 17, 2007 #7
    Yes it is. The proof: in any circle (M,r) draw a chord AB. Complete [itex]\triangle{AMB}[/itex].
    Now [itex]\angle{ABM}=\angle{BAM}=\alpha[/itex].
    Then [itex]\angle{AMB}=180-2\alpha[/itex] .
    Produce AM till it meets the circle in C, now [itex]\angle{BMC}=2\alpha[/itex].
    So arc BC is [itex]2\alpha[/itex].
  9. Jul 17, 2007 #8
    I'm not really sure what you proved edgo. You need to show that two chords that intersect on the circle have an angle uniquelly determined by the circle or equivalently that two pairs of such chords that cover the same arc have the same angle between them.
  10. Jul 17, 2007 #9
    I basically want someone to help me with the proof that, in the attachments (which are both the same), angle ACB is congruent to angles ADC if that is a circle and both angle span the same arc. It is a pretty basic proof that a lot of other proofs just take for granted but I think it is important to understand it. Sorry the drawing is kind of small--not really sure why.

    Thanks for your help guys.

    Attached Files:

    • POP.JPG
      File size:
      8.9 KB
    • POP.PNG
      File size:
      13 KB
    Last edited: Jul 17, 2007
  11. Jul 17, 2007 #10


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    The POP can be proven with vector algebra. If P lies in the chord AD, then you can let x be the (signed) ratio between the segments PA and DA. i.e.

    (D-A) = x (P-A)

    If you choose your coordinates s.t. the center of the circle is the origin, then you can explicitly compute the dot product of any pair of variables except P.P (which, of course, depends only on P). The power of P is just


    (or maybe the negative of that, depending on your definition).
  12. Jul 18, 2007 #11

    Gib Z

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    Ok Well Heres the proof for a related theorem, that the angle at the center is twice the angle at the circumference for angles standing on the same arc. Refer to the link and excuse my terrible circle, i couldn't find my compass: http://img362.imageshack.us/my.php?image=28954819hk1.jpg

    At the start of the proof, we only have Chords AB and AC, and radii AO and OC. Produce a line that goes through B and O to a point, P. Now let the angle ABO be [itex]\alpha[/itex], and angle CBO be [/itex]\beta[/itex], as in the diagram.

    Now notice that AO and OB are equal, since they are radii of a circle. Hence the triangle ABO is isosceles. Therefore the base angles are equal. From that we get angle BAO is also equal to [itex]\alpha[/itex].

    We know the angles in a triangle sum to 180 degrees, so we know that angle AOB = [itex] 180 -2\alpha[/itex]. And since the angle in a straight line is also 180, the angle AOP is [itex] 2\alpha[/itex].

    We repeat the exact same thing for the right side and get that angle COP is [itex] 2\beta[/itex].

    Angle ABC = [tex]\alpha + \beta[/tex]
    Angle AOC = [tex] 2 (\alpha + \beta)[/itex].

    ie Angle AOC = 2 (Angle ABC)

    Now to see that angles at the circumference standing on the same arc are equal, we just have to see that even though in the picture I drew B right up the top to look nice, we could have put it anywhere, such as the big bold X I drew. If I drew the angle AXC, the angle for that would still be [itex]\alpha + \beta[/itex], since the angle is still the same at the centre!

    I hope that cleared things up.
  13. Jul 18, 2007 #12
    I still think that I really gave you the proof that you asked for but please make that small drawing that I proposed. In [itex]\triangle{ABM}[/itex]the [itex]\angle{ABM}[/itex] = [itex]\angle{BAM}[/itex] because and only because that triangle has two sides AM and BM that are equal (r), this has nothing to do with arcs.The supplement of [itex]\angle{ABM}[/itex] is [itex]2\alpha[/itex] and this also has nothing to do with arcs. But, that supplement [itex]\angle{BMC}[/itex] is an angle with its top in M. It is just another theorem that such an angle is equal to the arc it is standing on, so in our case that arc BC must be [itex]2\alpha[/itex].
    That means that [itex]\angle{BAC}[/itex], which was [itex]\alpha[/itex], is half the arc it is standing on. Any other angle in this circle with the top on the circle and standing on arc BC is also half that arc BC.
    The answer to your question is one sentence long: all angles in a circle with their top on the circle and standing on the same arc are equal because they are all equal to half that arc they are standing on.
  14. Jul 18, 2007 #13
    Thanks Gib Z. That did clear things up.

    edgo, I think your proof does not hold for the general case since AC is a diameter (if I drew it correctly) and we want to prove the theorem for two chords neither of which need to be a diameter.
  15. Jul 18, 2007 #14
    There is no need for it but if you want so: just apply the same method to that second chord and call [itex]\alpha[/itex] now [itex]\beta[/itex]. As they share the same diameter AC both chords are now standing together on an arc [itex]2\alpha + 2\beta[/itex]. This implies that the angle between the two chords is now [itex]\alpha+\beta[/itex].
    I have the impression that you are sweating to lift a trunk that is empty.
  16. Jul 18, 2007 #15
    What if you draw the second chord between the first chord and the diameter? Then you can clearly see that the angle between the first and second chord is not alpha + beta because it must be smaller than alpha.
  17. Jul 18, 2007 #16
    [itex]\alpha-\beta[/itex] or [itex]\beta-\alpha[/itex] depending on which of the two is the biggest.
  18. Jul 18, 2007 #17
    Yeah--that completes the proof. Thanks edgo.
  19. Jul 18, 2007 #18
    [itex]\alpha - \beta[/itex] or [itex]\beta-\alpha[/itex] depending on which one is the biggest
  20. Jul 18, 2007 #19
    sorry I lost my reply on page 2. Y're welcome.
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