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Math Help - Geometry proof?

  1. #1
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    Geometry proof?

    I have this Geometry problem here which is a proof question.
    I am really unsure of where to start, but I have been told by the teacher that using congruent triangles might help, but does that mean I will need to draw in a line onto the diagram? I'm not quite sure where to start.
    Thanks.
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let \widehat{ABC}=\alpha, \widehat{BAC}=2\alpha, \widehat{ACD}=\widehat{DCB}=\beta

    In the triangle BDC we have CD=\frac{BC\sin\alpha}{\sin(\alpha+\beta)}

    In the triangle ADC we have CD=\frac{AD\sin 2\alpha}{\sin\beta}

    Then \frac{BC\sin\alpha}{\sin(\alpha+\beta)}=\frac{2AD\  sin\alpha\cos\alpha}{\sin\beta}\Rightarrow BC\sin\beta=2AD\sin(\alpha+\beta)\cos\alpha\Righta  rrow

    \Rightarrow BC\sin\beta=AD[\sin(2\alpha+\beta)+\sin\beta]\Rightarrow BC=\frac{AD\sin(2\alpha+\beta)}{\sin\beta}+AD=

    =\frac{AD\sin(\alpha+\beta)}{\sin\beta}+AD=AC+AD
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  3. #3
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    Quote Originally Posted by f1f2f3 View Post
    I have this Geometry problem here which is a proof question.
    I am really unsure of where to start, but I have been told by the teacher that using congruent triangles might help, but does that mean I will need to draw in a line onto the diagram? I'm not quite sure where to start.
    Thanks.
    I wonder if this is the construction that the teacher had in mind. Let E be a point on BC such that angle EDC = angle ADC. Then see if you can prove these facts:
    1. Triangles DEC, DAC are congruent.
    2. Angle DEC = angle DAC = 2 angle DBE.
    3. Triangle BDE is isosceles.
    4. BE = DE = DA.
    5. BC = BE + EC = AD + AC.
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  4. #4
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    Sorry, I'm not quite sure how I'm supposed to prove the first step. How are triangles DEC and DAC congruent?
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  5. #5
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    Quote Originally Posted by f1f2f3 View Post
    Sorry, I'm not quite sure how I'm supposed to prove the first step. How are triangles DEC and DAC congruent?
    The side CD is common to both triangles.
    Angle EDC = angle ADC (by construction).
    Angle ECD = angle ACD (because CD bisects angle ACB).
    Therefore the triangles are congruent (one is a mirror image of the other).
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  6. #6
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    Also, how do I know that triangle BDE is isosceles?
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  7. #7
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    Quote Originally Posted by f1f2f3 View Post
    Also, how do I know that triangle BDE is isosceles?
    The external angle DEC is the sum of the internal angles DBE + BDE. But we know that angle DEC = angle BAC, which we are told is equal to twice angle DBE. So angle DBE is half of angle DEC, and therefore so is angle BDE. Thus the angles DBE and BDE are equal.
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