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Thread: Circle Geometry Extension 1 Help

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    Circle Geometry Extension 1 Help

    1. Let A, B, C, D, and E be five points in order around a circle with centre O, and let AOE be a diameter. Prove that <ABC + <CDE = 270◦.

    2a) Prove that if two chords of a circle bisect each other, then they are both diameters.
    b) Prove that if the chords AB and PQ intersect at M and MA = MP, then MB = MQ, BP = AQ and AP ∥ QB.


    Can someone solve these questions? Thanks

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    Re: Circle Geometry Extension 1 Help

    Q1

    Draw diameter AE. Pick any arbitrary points B, C and D between A and E on, say, the upper semi-circle (I used clockwise order). Join B and E. Then angle EBC + angle CDE = 180 (BCDE being cyclic). Angle ABE = 90 (angle in a semi-circle)

    Adding, angle ABC + angle CDE = 270.
    Last edited by Drongoski; 20 Aug 2018 at 9:15 PM.
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    Re: Circle Geometry Extension 1 Help

    Q2

    a) Let chord AB bisect chord CD (so A, C, B and D are in a clockwise order)

    Then we can use following properties to prove AB and CD are diameters:

    i) if AB bisects CD, then ACBD is a parallelogram.

    ii) since ACBD is cyclic, opp angles are supplementary
    .: angle ACB + angle BDA = 180

    iii) but ACBD is a //gram so that opp angles are equal

    iv) .: angle ACB = angle BDA = 90

    v) in cyclic quad ACBD, since angle ACB = 90, then AB must be a diameter (angle in a semi-circle = 90)

    vi) .: ACBD must be a rectangle and .: angle CAD = 90 = angle in a semi-circle & .: CD is a diameter
    Last edited by Drongoski; 21 Aug 2018 at 8:50 AM.
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    Re: Circle Geometry Extension 1 Help

    Q2

    b) it can be easily shown triangle AMP ||| triangle BMQ

    Since AM = MP, AMP is isosceles and .: BMQ is also isosceles and .: MB = MQ so that the 4 base angles are all equal.

    Since ang APM = ang PQB and they are alternate angles, AP//BQ

    Now triangles BAQ & QPB are congruent, noting ang BAQ = ang QPB and AB = PQ and BQ = QB

    .: corresponding sides AQ = BP
    Last edited by Drongoski; 23 Aug 2018 at 1:58 PM.
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