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Math Help - part one of the hard question trigiolgy

  1. #1
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    part one of the hard question trigiolgy

    P, Q are points on the side AC, AB of triangle ABC such that
    angBPC = ang BQC; BP cuts CQ at K; X, Y are points on AC, AB such that KXAY is a parallelogram. Prove that (i) AX.XC = AY.YB (ii) the centre of the circle ABC is equidistant from X and Y

    I hav already solve (i) but just to typed if it is of any help to (ii)
    PS sorry but i dont know how to draw diagrams on computer
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  2. #2
    Senior Member OReilly's Avatar
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    Quote Originally Posted by calculus_jy View Post
    P, Q are points on the side AC, AB of triangle ABC such that
    angBPC = ang BQC; BP cuts CQ at K; X, Y are points on AC, AB such that KXAY is a parallelogram. Prove that (i) AX.XC = AY.YB (ii) the centre of the circle ABC is equidistant from X and Y

    I hav already solve (i) but just to typed if it is of any help to (ii)
    PS sorry but i dont know how to draw diagrams on computer
    What does this means AX.XC = AY.YB !?
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  3. #3
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    Speification

    AX.XC = AY.YB means the length of AX multiple by the length of XC equals the length of AY multiple by the length of YB
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  4. #4
    Senior Member OReilly's Avatar
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    Quote Originally Posted by calculus_jy View Post
    AX.XC = AY.YB means the length of AX multiple by the length of XC equals the length of AY multiple by the length of YB
    Can you post that proof?

    I have uploaded diagram.
    Attached Thumbnails Attached Thumbnails part one of the hard question trigiolgy-untitled.jpg  
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  5. #5
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    proof

    since angBQC = angCPB (given) and are on the same side of BC, the pts BQPC are concylic (equal angles subtended on same side of a line)

    Therefore, angYBK=angQBK = ang XCK=angXCK (angles subtended on same arc QP)

    In parallelogram AYKX, angAYK = angAXK (opp ang of parallelgram)
    Therefore angBYK = 180 deg - angAYK = 180 deg - angAXK = angCXK

    Then it can be proven with angBYK = angCXK and angYBK = angXCK that the triangles YKB and XKC are equiangular

    thus YK / XK = YB / XC (sides proportional in eqiangular tri)

    YK . XC = YB . XK

    But YK = AX (opp sides of parallelogram) and similiarly AY= XK

    AX. XC = AY. YB

    Hope that helps to do part (ii) because i really hav no idea
    Last edited by calculus_jy; June 17th 2007 at 11:03 PM. Reason: Typing error
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  6. #6
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    Quote Originally Posted by calculus_jy View Post
    since angBQC = angCKP and are on the same side of BC, the pts BQPC are concylic (equal angles subtended on same side of a line)
    Angles BQC and CKP are not equal, I don't know how did you get that?
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  7. #7
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    Quote Originally Posted by calculus_jy View Post
    P, Q are points on the side AC, AB of triangle ABC such that
    angBPC = ang BQC; BP cuts CQ at K; X, Y are points on AC, AB such that KXAY is a parallelogram. Prove that (i) AX.XC = AY.YB (ii) the centre of the circle ABC is equidistant from X and Y
    I hav already solve (i) but just to typed if it is of any help to (ii)
    PS sorry but i dont know how to draw diagrams on computer
    Hello,

    I've attached a sketch of this problem.

    1. The angles \angle(BQC) and \angle(BPC) are equal and they are subtended over the same side of the triangle thus the points B, Q, P, C must lie on a circle with M_2 as centre (green).

    2. Therefore you can use the intersecting chord theorem here, using BP and CQ as chords and K as intersection.

    3. The triangles YQK and PXK are similar.

    4. With the green circle and the secants AC and AB you can use the secant-secant-theorem.

    5. The two centres M_1 and M_2 must lie on a perpendicular bisector of BC. And M_1 must lie on a perpendicular bisector of XY.

    But: I haven't got a clue how to put all those facts together to proof the statement of your problem.
    Attached Thumbnails Attached Thumbnails part one of the hard question trigiolgy-parallgram_in_dreick.gif  
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  8. #8
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    Thanks

    regarding to oreilly, it was a typing mistake that i made.

    earthboth, i dont get what that intersecting chord theorem in 2 is used for, also i dont get what the secant-secant theorm is and used for either,
    however, it will be nice if u can provide the proof that
    must lie on a perpendicular bisector of XY as this is the prove that the centre is eqidistant from XY

    Thank you so much for both ur help
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