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Basic geometry theorems

  1. Jan 19, 2008 #1


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    My problem is how to understand this problem. I am confident that I can solve it, but I need to know what they mean in the problem.

    1. The problem statement, all variables and given/known data

    "[PQ] is a chord of a circle. R lies on the major arc of the circle."
    I understand this.

    "Tangents are drawn through P and through Q."
    What does this mean? I would say that they meant that there are tangents touching the circle in P and Q.

    "From R, perpendicuars [PA], [PB] and [PC] are drawn to the tangent at A, the tangent at B and [AB] respectively."
    I'm totally lost. What do they mean with this?

    "Prove that [tex]RA.RB=RC^2[/tex]"
    This is the problem I am to prove, when I understand the problem.

    2. Relevant equations

    Basic geometry theorems

    3. The attempt at a solution

    Just need some help with understanding the problem.
    Last edited: Jan 19, 2008
  2. jcsd
  3. Jan 19, 2008 #2
    Did this come with a diagram, by any chance?
  4. Jan 19, 2008 #3


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    Nope. In problems with a diagram, the text "In the given figure" appears.
  5. Jan 19, 2008 #4
    Something just doesn't make sense to me.

    Since the perpendiculars are labeled PA, PB, and PC, I'm assuming that they all originate from point P. So how could they be drawn from R? I'm also assuming that points A and B lie on the circle. If so, the lines perpendicular to them would have to go through the center, and it should not be possible to draw lines perpendicular to two different tangents from the same point on the circle, whether R or P.

    Only one perpendicular to a tangent can be drawn from R, and that's the diameter through R. The same goes for P, since P also lies on the circle.

    I can't help but think there's some missing information here. I'm sorry if this just adds to the confusion, but I just can't see how any figure could possibly be drawn to the specifications here.
  6. Jan 19, 2008 #5


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    Excactly! :/

    I can't either see the figure. I guess I could skip this excersise with good conscience then :)


    (I think this chapter in the book was hastily written together)
  7. Jan 19, 2008 #6


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  8. Jan 20, 2008 #7


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    Thanks for the reply! It is no way that C is on AB though. RC squared cannot be the same as the product of RA-RB. I drew the figure in geogebra, and if we place the perpendicular line from R to the segment PQ, then the thing we are to prove seem very correct. I try to move the point R around on the figure, and RA.RB-RC^2 = 0 for all points R on the circle. I will prove this instead. If C were on AB, it would not be right for any R, as RC is smaller than both RA and RB for any R (on my figure).

    Proof: (See attached picture for helping figure)

    Let the tangents in P and Q intersect in M. Let [tex]\angle MPQ= \alpha[/tex]. As [tex]\Delta MPQ[/tex] is an isosceles, [tex]\angle PQM = \alpha[/tex]. By the "angle between a chord and a tangent" theorem, [tex]\angle PRQ = \alpha[/tex].
    Let [tex]\angle ARP = \beta \Rightarrow \angle RPA = 90-\beta \Rightarrow \angle RPC = 90+\beta - \alpha[/tex].
    Now in [tex]\Delta PRQ[/tex], [tex]\angle PRQ - \angle CRQ + \angle RPC +\angle PCR = 180 \Rightarrow \angle CRQ = \beta[/tex].

    Now we have that [tex]\Delta ARP[/tex] is similiar to [tex]\Delta CRQ[/tex].
    Then [tex]\frac{RA}{RP}=\frac{RC}{RQ} \Rightarrow \frac{RC}{RA}=\frac{RQ}{RP}[/tex].

    Now as [tex]\angle CQR = 90 - \beta, \angle RQB=90+\beta - \alpha \Rightarrow \Delta PCR[/tex] is similar to [tex]\Delta QBR[/tex].
    Then [tex]\frac{RP}{RC}=\frac{RQ}{RB} \Rightarrow \frac{RQ}{RP}=\frac{RB}{RC}[/tex] and hence is [tex]\frac{RC}{RA}=\frac{RB}{RC} \Rightarrow RA.RB=RC^2[/tex]


    Does it seem ok?

    I believe that this is what the problem meant.

    Attached Files:

    Last edited: Jan 20, 2008
  9. Jan 20, 2008 #8
    I think this was a typo. You meant [tex]\angle PRQ = \alpha[/tex], right?
    Looks good to me! Well done.

    I had also initially thought that the point C was supposed to be on the the line AB, but your proof shows that it must be as you interpreted it.
  10. Jan 20, 2008 #9


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    that problem had a couple of typos.

    yeah, I meant that.
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