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Math Help - Arc over right angle proof

  1. #1
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    Arc over right angle proof

    On two sides of right angle with vertex in point O we choose points A and B, |OA|=|OB|. On arc AB with center in O we choose point P. Through that point we draw line parallel with AB which intersects OA in point C, and OB in point D. Prove that |PC|2+|PD|2=|AB|2.

    I hope you understand my question. I find a lot of similar and right triangles but always get stuck.
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  2. #2
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    Re: Arc over right angle proof

    Hello, kicma!

    On two sides of right angle O we choose points A and B, |OA|=|OB|.
    On arc AB with center in O we choose point P.
    Through that point we draw line parallel with AB
    which intersects OA in point C, and OB in point D.
    Prove that: |PC|2+|PD|2=|AB|2.

    Code:
          |
        D *
          | *
          |   *
        B *     *
          | *     *  P
          |   *     o
          |     * .   *
        r |     . *     *
          |   .     *     *
          | . θ   45o *  45o *
    
        - + - - - - - - * - - -* - - -
          |O     r      A      C
    We have quarter circle AOB.
    . . r = AO = OB,\;\angle BAO = \angle ABO = 45^o.
    Note that: AB \,=\, \sqrt{2}r \quad\Rightarrow\quad \overline{AB}^2 \,=\,2r^2

    P is on arc AB.
    CPD \parallel AB.\;\;\angle DCO = \angle CDO = 45^o.
    Let \theta = \angle POC.

    We will use the Law of Sines.

    \text{In }\Delta OPC\!:\;\frac{PC}{\sin\theta} = \frac{r}{\sin45^o} \quad\Rightarrow\quad PC \,=\,\frac{r\sin\theta}{\sin45^o}

    \text{In }\Delta DOP\!:\;\frac{PD}{\sin(90-\theta)} = \frac{r}{\sin45^o} \quad\Rightarrow\quad PD \,=\,\frac{r\cos\theta}{\sin45^o}

    \text{Hence: }\:\overline{PC}^2 + \overline{PD}^2 \;=\;\frac{r^2\sin^2\theta}{\frac{1}{2}} + \frac{r^2\cos^2\theta}{\frac{1}{2}}

    . . . . . \overline{PC}^2 + \overline{PD}^2 \;=\;2r^2\sin^2\theta + 2r^2\cos^2\theta

    . . . . . \overline{PC}^2 + \overline{PD}^2 \;=\;2r^2(\sin^2\theta + \cos^2\theta)


    Therefore: . \overline{PC}^2 + \overline{PD}^2 \;=\;2r^2\;=\;\overline{AB}^2
    Thanks from kicma
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  3. #3
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    Re: Arc over right angle proof

    I forgot to mention I'm not allowed to use Law of Sines/Cosines. There should be a way to prove this without it if you can help.
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