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Thread: circle/ iscosceles triangle

  1. #1
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    circle/ iscosceles triangle

    In the given fig., ABC is an isosceles triangle in which AB = AC. A circle through B touches the side AC at D and it intersects the side AB at P. If D is the mid-point of AC, prove AB = 4AP.
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  2. #2
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    Hello, snigdha!

    I used a (sort of) well-known theorem:

    If a tangent and a secant are drawn to a circle from an external point,
    the tangent is the mean proportional of the secant and the external segment of the secant.


    $\displaystyle ABC$ is an isosceles triangle wherte $\displaystyle AB = AC.$
    A circle through $\displaystyle B$ touches side $\displaystyle AC$ at $\displaystyle D$, and intersects side $\displaystyle AB$ at $\displaystyle P$.
    If $\displaystyle D$ is the midpoint of $\displaystyle AC$, prove that: .$\displaystyle AB \,=\, 4\!\cdot\!AP$
    Code:
                  B
                  ♥ o o
              o  *  *     o
            o   *     *     o
           o   *        *    o
              *           *
          o  *              * o
          o *                 ♥ P
          o*                  o *
          *                       *
         * o                 o      *
        *   o               o         *
       *      o           o             *
      ♥   *   *   o ♥ o   *   *   *   *   ♥ A
                    D

    We are given: .$\displaystyle AB \,=\,AC$

    $\displaystyle D$ is the midpoint of $\displaystyle AC\!:\;\;AD \,=\,\tfrac{1}{2}AB \quad\Rightarrow\quad AD^2 \,=\,\tfrac{1}{4}AB^2$ .[1]


    From the theorem, we have: .$\displaystyle AD^2 \:=\:AB\cdot AP$ .[2]


    Substitute [1] into [2]: .$\displaystyle \tfrac{1}{4}AB^2 \:=\:AP\cdot AB \quad\Rightarrow\quad AB^2 \:=\:4\!\cdot\!AP\!\cdot\!AB$

    . . Therefore: .$\displaystyle AB \:=\:4\!\cdot\!AP$

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