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Thread: geometry (3)

  1. #1
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    geometry (3)

    In the diagram , AOB is a diameter of the circle with centre O . ACX , BDY and DEC are tangents to the circle at A , B and E respectively . Show that

    (1) angle COD =90

    (2) OC . OD = 1/2 AB . CD
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  2. #2
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    Hello thereddevils
    Quote Originally Posted by thereddevils View Post
    In the diagram , AOB is a diameter of the circle with centre O . ACX , BDY and DEC are tangents to the circle at A , B and E respectively . Show that

    (1) angle COD =90

    (2) OC . OD = 1/2 AB . CD
    Part (1) is like your previous question: Geometry (1). Here's the outline of the proof. I'll leave the details to you. Let $\displaystyle \angle ACO = x, \angle EDO = y$.

    Then $\displaystyle \angle ECO = x$ and $\displaystyle \angle BDO = y$ (congruent $\displaystyle \triangle$'s)

    Then look at the quadrilateral $\displaystyle ACDB$. Can you see why $\displaystyle 2x+2y = 180^o$? Hence $\displaystyle x+y=90^o$. Then look at $\displaystyle \triangle COD$, and you're there.

    Part (2) Let $\displaystyle AC = CE = p,\,ED=BD = q,\, AO=EO=BO =r$.

    Then look at the areas:

    $\displaystyle OC\cdot OD = 2\triangle OCD$
    $\displaystyle = 2(\triangle OCE + \triangle ODE)$

    $\displaystyle =2(\tfrac12pr + \tfrac12qr)$

    $\displaystyle =(p+q)r$

    $\displaystyle = \tfrac12CD\cdot AB$
    Grandad
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