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Math Help - Circles

  1. #1
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    Circles

    Let  \bold{r} be a point on a circle with diameter from  - \bold{a} to  \bold{a} (so that  |\bold{r}| = |\bold{a}| ). Draw the chord through  \bold{r} perpendicular to  \bold{a} (the length and direction of this chord is  2 \bold{r}_{a \perp} ). Let  \bold{t} = \bold{r}_{a} be the point of intersection of this chord with the diameter. Prove that the square of half the length of the chord is the product of the distance from  -\bold{a} to  \bold{t} times the distance from  \bold{t} to  \bold{a} :  |\bold{r}_{a \perp}|^{2} = (\bold{t} + \bold{a}) \cdot (\bold{a} - \bold{t}) .
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  2. #2
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    So I got  \hat{\bold{a}} = \frac{\bold{a}}{|\bold{a}|}

    and  \bold{t} = \bold{r} \cdot \hat{\bold{a}}

    and  \bold{r}_{a \perp} = \bold{t} - \bold{r} .

    Then  \bold{a} + \bold{r} gives one hypotenuse. And so  |\bold{a} + \bold{r}|^{2} = |\bold{r}_{a}|^{2} + |\bold{a}+\bold{t}|^{2} .

    So  |\bold{r} - \bold{a}|^{2} = |\bold{r}_{a}|^{2} + |\bold{a}-\bold{t}|^{2} .
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  3. #3
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    Quote Originally Posted by heathrowjohnny View Post
    Let  \bold{r} be a point on a circle with diameter from  - \bold{a} to  \bold{a} (so that  |\bold{r}| = |\bold{a}| ). Draw the chord through  \bold{r} perpendicular to  \bold{a} (the length and direction of this chord is  2 \bold{r}_{a \perp} ). Let  \bold{t} = \bold{r}_{a} be the point of intersection of this chord with the diameter. Prove that the square of half the length of the chord is the product of the distance from  -\bold{a} to  \bold{t} times the distance from  \bold{t} to  \bold{a} :  |\bold{r}_{a \perp}|^{2} = (\bold{t} + \bold{a}) \cdot (\bold{a} - \bold{t}) .
    Hello,

    you are dealing with a right triangle (see attachment)

    If the legs of a right triangle are a and b and the hypotenuse is c then you know:

    a^2+b^2 = c^2 ....... and c = p+q. That means c^2 = p^2+2pq+q^2

    Therefore:

    a^2+b^2 = p^2+2pq+q^2

    According to the drawing you have:

    a^2 = h^2 + p^2 ....... and b^2 = h^2 +q^2

    So finally you get:

    h^2 + p^2 + h^2 +q^2 = p^2+2pq+q^2~\iff~ 2h^2 = 2pq~\iff~\boxed{h^2 = pq}
    Attached Thumbnails Attached Thumbnails Circles-euklid_theorem.gif  
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