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Math Help - triangle + equations

  1. #1
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    triangle + equations

    This is my problem: We have a triangle ABC, that has a right angle at the vertex C. First we draw the height v from the vertex C, the heigh v intersects the line AB in point D. Now we draw 2 lines from point D, x and y, one that is parallel with line AC, the other parallel with line CB. Now I have to prove that c*x*y=v^3
    this is the photo:

    I know the problem is relatively simple: (using the pythagorean theorem multiple times )it just requires equation solving but I'm really lazy/stupid so if I could get any hints or help that would be really great...
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  2. #2
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    Lexington, MA (USA)
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    Hello, snezic!

    If we're allowed to use trig, I have a solution.
    (And I'm sure someone will find a shorter or neater one.)


    We have a triangle ABC, that has a right angle at the vertex C.
    We draw the altitude v = CD.
    Now we draw two lines from point D:
    . .  x = DE parallel to AC,\;\text{ and }\;y = DF parallel to BC.

    Prove that: . cxy\:=\:v^3


    Code:
        B o
          |  *
          |     *
          |     θ  *   D
        E o - - - - - o
          |     x    /:  *
          |         / :     *
          |        /  :        *
          |       / θ :           *
          |      /    :              *
         y|     /v    :y                *
          |    /      :                    *
          | θ /       :                       *
          |  /        :                          *
          | /         :                             *
          |/          :                             θ  *
        C o - - - - - o - - - - - - - - - - - - - - - - - o A
                x     F

    Note that all the right triangles are similar.


    \begin{array}{ccccccccc}\text{In }\Delta DFA\!: &\csc\theta &=& \dfrac{DA}{y} &\Rightarrow& DA &=& y\csc\theta & [1] \\ \\[-3mm]<br />
\text{In }\Delta BED\!: & \sec\theta &=&\dfrac{BD}{x}  & \Rightarrow& BD &=& x\sec\theta & [2] \end{array}

    . . \begin{array}{cccccc}\text{In }\Delta DFC\!: & \csc\theta &=& \dfrac{v}{x} & [3] \\ \\[-3mm]<br />
\text{In }\Delta DEC\!: & \sec\theta &=&\dfrac{v}{y} & [4] \end{array}

    Substitute [3] and [4] into [1] and [2]: . \begin{array}{cccccc}DA &=& y\cdot \dfrac{v}{x} &=& \dfrac{vy}{x} & [5]\\ \\[-3mm] <br />
BD &=& x\cdot\dfrac{v}{y} &=& \dfrac{vx}{y} & [6] \end{array}

    \text{Add [5] and [6]: }\;\underbrace{DA + BD}_{\text{This is side }c} \;=\;\frac{vy}{x} + \frac{vx}{y}\;=\;v\left(\frac{y}{x} + \frac{x}{y}\right)

    \text{And we have: }\;c \;=\;v\cdot\frac{\overbrace{x^2+y^2}^{\text{This is }v^2}}{xy} \quad\Rightarrow\quad c \;=\;v\cdot\frac{v^2}{xy}


    . . Therefore: . cxy \;=\;v^3

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  3. #3
    Newbie
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    Thank you very much...
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