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Help with Geometry Problems

  1. Aug 4, 2004 #1


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    Hi all. I'm looking at some old Maths exam papers at the moment and there is one particular type of question that has me stumped. They look just like basic geometry questions of the type I can normally do, but I think I'm missing something in these particular problems as I'm getting stuck.

    I've uploaded an example of the type under consideration and would really appreciate a hand getting started with it. Ideally if someone could just explain how to get the answer in part (i) I'd like to see if I can take it from there.


    Attached Files:

    • q6b.gif
      File size:
      7.8 KB
  2. jcsd
  3. Aug 4, 2004 #2
    Show us what you've done so far.

    Two triangles are 'similar' if thier corresponding angles are equal and corresponding line segments proportional.

    Does that help?
  4. Aug 4, 2004 #3
    so far I got ATF = ATE (common angle)
    AT = AT (common side)
    AFE = FAT + FTA (Exterior Angle Theorem)
    Im not quite shure where to go from now. I hope this helps.
  5. Aug 4, 2004 #4


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    Yeah I'm ok with the stuff regarding similar triangles, corresponding angles, alternate angles, tangents and so on. I used all the stuff I know but kept coming up just short of a solution, like there was one peice of info or property that I wasn't using.

    The point I got stuck at was this. I was trying to prove that angles FAT = FEA, that would get me part(i) ok. But after hitting it with everything I knew the closest that I could get was FAT + FTB = FEA + EAF.

    I knew that if I could somehow prove that FTB = EAF then I'd have it, but I couldn't seem to do it and suspected I was missing something.

    I'm thinking that there must be some special property of a triangle that is inscribed upon a circle that I dont know about but need to use. I've got a hunch that I've just thought of, maybe someone can confirm that this is indeed a property of such a triangle.

    Hunch : If the two vertices of a triangle are fixed points on a circle, and the third vertex of the triangle is free to move upon the circle, then the angle subtended by the moving vertex is a constant (so long as it stays on the circle and doesn't cross either of the two fixed vertices).

    Is the above a fact?
  6. Aug 5, 2004 #5
    Hi uart,

    Sorry for the short reply ... I'll have a proper look later.

    I had the attached gif from a previous post on a similar (no pun intended) problem.

    This shows the 'hunch' you had is good.

    As I say, I'll have a proper look later.


    Attached Files:

  7. Aug 5, 2004 #6


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    Thanks pnaj, your graphic describes exactly what I was thinking of. Now I'll have to try and find a proof for it.

    BTW, the one above result also leads me directly to two other results that also help in some of these problems. These are :

    1. The two opposing triangles produced by two intersecting cords on a circle are similar.

    2. The sum of opposite angles in a quadrilateral inscribed on a circle is 180 degrees.

    Anyway, together those results greatly help me in this and some similar problems that I previously couldn't do. :)
    Last edited: Aug 5, 2004
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