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Math Help - deductive geometry (1)

  1. #1
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    deductive geometry (1)

    Points A and B are on the side XY of triangle XYZ with XA=AB=BY . Points C and D are on the sides YZ and XZ respectively such that ABCD is a rhombus . Prove that angle XYZ is 90 degree .
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  2. #2
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    Hello, thereddevils!

    Please check the wording of the problem.
    As stated, the conclusion cannot be true.


    Points A and B are on the side XY of \Delta XYZ with XA=AB=BY.
    Points C and D are on the sides YZ and XZ resp. such that ABCD is a rhombus.
    Prove that \angle XYZ \,=\,90^o.
    Code:
        X o
          |\
          | \
          |  \
          |   \
          |    \
        A o     \
          | *    \
          |   *   \
          |     *  \
          |       * \
          |         *\
        B o           o D
          | *         |\
          |   *       | \
          |     *     |  \
          |       *   |   \
          |         * |    \
        Y o - - - - - o - - o Z
                      C
    Suppose \angle XYZ = 90^o.


    We are given: . BY = AB

    Since ABCD is a rhombus: . BC = AB.

    Hence: . BY = BC.

    Then we have right triangle BYC in which side BY
    . . and hypotenuse BC are equal. ??

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  3. #3
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    not sure.. perhaps they ask u to prove angle XZY is 90
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  4. #4
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    Quote Originally Posted by Soroban View Post
    Hello, thereddevils!

    Please check the wording of the problem.
    As stated, the conclusion cannot be true.

    Code:
        X o
          |\
          | \
          |  \
          |   \
          |    \
        A o     \
          | *    \
          |   *   \
          |     *  \
          |       * \
          |         *\
        B o           o D
          | *         |\
          |   *       | \
          |     *     |  \
          |       *   |   \
          |         * |    \
        Y o - - - - - o - - o Z
                      C
    Suppose \angle XYZ = 90^o.


    We are given: . BY = AB

    Since ABCD is a rhombus: . BC = AB.

    Hence: . BY = BC.

    Then we have right triangle BYC in which side BY
    . . and hypotenuse BC are equal. ??
    Sorry Soroban , it should be prove that angle XZY = 90 .
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  5. #5
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    Hello, thereddevils!

    It should be: prove that \angle XZY \,=\, 90^o . . . . Ah, got it!

    Code:
        X o
          |\
          | \
          |  \
          |   \
          |    \
          |     o A
          |    / \
          |   / θ \
          |  /     \
          | /       \
          |/         \
        D o           o B
          |\         / \
          | \       / θ \
          |  \     /     \
          |   \   /       \
          |    \ /         \
        Z o - - o - - - - - o Y
                C

    We have: . XA \,=\,AB \,=\,BY \,=\,BC\,=\,CD\,=\,DA

    Let \theta \,=\,\angle CBY


    Since \Delta CBY is isosceles, \angle BCY = \angle BYC.
    Then: . \angle BYC + \angle BCY + \theta \:=\:180^o \quad\Rightarrow\quad \boxed{\angle BYC \,=\,90^o - \tfrac{\theta}{2}}


    Since AD \parallel BC\!:\;\angle BAD = \theta \quad\Rightarrow\quad \angle XAD \,=\,180^o-\theta

    Since \Delta XAD is isosceles: \angle AXD \,=\,\angle ADX
    Then: . \angle AXD + \angle ADX + (180^o -\theta)  \:=\:180^o \quad\Rightarrow\quad \boxed{\angle AXD \,=\,\tfrac{\theta}{2}}


    In \Delta XYZ\!:\;\;\angle X \,=\, \tfrac{\theta}{2},\;\;\angle Y \,=\, 90^o-\tfrac{\theta}{2} . . . They are complementary!

    Therefore: . \angle Z \:=\:90^o

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