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Math Help - deductive geometry (2)

  1. #1
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    deductive geometry (2)

    Vertices B and C of the triangle ABC lie on the circumference of a circle . AB and AC cut the circumference of the circle at X and Y respectively . Show that angle CBX +angle CYZ =180

    If AB=AC , show that BC is parallel to XY .
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  2. #2
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    Quote Originally Posted by thereddevils View Post
    Vertices B and C of the triangle ABC lie on the circumference of a circle . AB and AC cut the circumference of the circle at X and Y respectively . Show that angle CBX +angle CYX =180

    If AB=AC , show that BC is parallel to XY .



    let the centre of the circle be O

    let angle COX(major) be 2x
    angle CBX = 0.5 angle COX(major) = x
    (angle of circumference is 1/2 of the angle at centre)

    angle COX ( minor) + angle COX(major) = 360 (angel at a point)
    thereofre angle COX (minor) = 360 -2x

    angle CYX = 0.5 angle COX(minor) = 180 -x
    (angle of circumference is 1/2 of the angle at centre)


    so angle CBX + angle CYX = x + 180 -x = 180 (proven)
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  3. #3
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    Hello, thereddevils!

    The first part has theorem . . .


    Vertices B and C of \Delta ABC lie on the circumference of a circle.

    AB and AC cut the circumference of the circle at X and Y respectively.

    Show that: . \angle CBX + \angle CYX \:=\:180^o
    Theorem:
    . . If a quadrilateral is inscribed in a circle, the opposite angles are supplementary.


    Code:
               X    * * *
                o           * 
              *               *
             *                 o Y
    
            *                   *
            *                   *
            *                   o C
    
             o                 *
           B  *               *
                *           *
                    * * *

    Draw chords XY,\: YC,\: CB,\: BX.

    . . \begin{array}{cccccccc}\angle Y \:=\:\frac{1}{2}\text{arc}(XBC) & [1] \\ \\[-4mm] \angle B \:=\:\frac{1}{2}\text{arc}(XYC) & [2] \end{array}

    Add [1] and [2]: . \angle Y + \angle B \;=\;\tfrac{1}{2}\text{arc}(XBC) + \tfrac{1}{2}\text{arc}(XYC)

    . . . . . . . . . . . . \angle Y + \angle B \;=\; \tfrac{1}{2}\bigg[\text{arc}(XBC) + \text{arc}(XYC)\bigg] \;=\;\tfrac{1}{2}(360^o)

    . . . . . . . . . . . . \angle Y + \angle B \;=\;180^o

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  4. #4
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    (1) simply apply the theorem for inscribed quadrilateral....

    or

    (2) draw two radii XO and CO where O is the centre of circle. There is a theorem (forgotten its name) saying that reflex angle XOC = angle XYC x 2, and obtuse angle XOC = angle XBC x 2.
    since the sum of the two angles XOC equals 360 degrees:
    angle XYC x 2 + angle XBC x 2 = 360 degrees, hence
    angle XYC + angle XBC = 180 degrees
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