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Geometry proof

  1. Dec 12, 2004 #1
    I finished this proof that took about a whole page and I was just wondering if there were an easier way of doing things that would take less work. here's the problem.
    PQ and PR are two chords of a circle with centre O. OT is perpendicular to PQ and OS is perpendicular to PR. IF OT = OS prove that T,S,R and Q are concyclic.
    I am going to go through my steps without proof cause it would take forever. I just want to get an idea if I am doing it the easiest way.

    It would probably help if you drew this out.

    First I proved QT = RS
    then QTO is congruent to RSO
    Then you can prove TS is parallel to QR
    from there i can conclude that <QSR = <QTR
    from there can I conlude that they are concyclic by equal angles in a segment?

    Edit: I realized that I didn't need to prove parallel lines. It really doesn't help the question in any way.
  2. jcsd
  3. Dec 12, 2004 #2
    Ya, as I look at it, I dont think this proof works. I think I assumed O was on QR which it doesn't have to be. Now I really need help !!
  4. Dec 13, 2004 #3
    T and S are not defined.
    If you don't tell a little bit more about them, is gonna be impossible to prove anything.
  5. Dec 15, 2004 #4
    I suppose that T lies on PQ and S lies on PR
    This two equalities make the situation simple

    At first we may notice that
    If AB is a chord of a circle with centre O then the triangle OAB is an isoskeles, so the perpendicular from O to AB is at once
    -perpendicular bisector of AB
    -angle bisector of AOB

    The right triangles OPS,OPT are congruent (OS=OT, OP common)
    So we get PS=PT
    Then, it's easy to prove that the four triangles OPS,ORS,OPT,OQT are congruent
    (actually there is a symmetry trough OP)

    The way I use is, to prove that all perpendicular bisectors of RS,RQ,QT intersect each other at the point K.
    Then KR=KS,KR=KQ,KQ=KT (so, the point K is the center of the circle we are looking for)

    You can easy show that the perpendicular bisector on RQ is the line OP (using congruent triangles)

    About the two other,
    Take the midpoint A of RS, and respectively the midpoint B of TQ

    bring the perpendicular bisector of RS, it intersects OP at the point K (theydo intersect since AK is parallel to SO and SO intersects PR)

    You have the right triangle AKP,

    from K bring the perpendicular line to PQ, which intersects PQ at the point B
    Now, you have another right triangle BKP
    But their angles BPK and APK are equal => Triangles AKP,BKP are similar
    If you use that KP is common, you get that AKP and BKP triangles are congruent

    So KA=KB
    and you can use it to prove KR=KS=KT=KQ

    I hope that helps

    You can simplify further the proof if you use symmetry
    Everything over OP is symmetric to everything on the other side of OP
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