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Thread: Deductive geometry, tangents

  1. #1
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    Deductive geometry, tangents

    Hi, I have a problem which I am having trouble solving.

    Let $\displaystyle ABC$ be a right-angled triangle. A circle $\displaystyle T$ having side $\displaystyle AC$ as its diameter meets hypotenuse $\displaystyle AB$ at point $\displaystyle E$. A tangent line to $\displaystyle T$ at point $\displaystyle E$ meets side $\displaystyle BC$ at point $\displaystyle D$. Prove that triangle $\displaystyle BDE$ is isosceles.

    Please help, BG
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  2. #2
    MHF Contributor red_dog's Avatar
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    Let T be the center of the circle.

    Quadrilater CTED is cyclic. Then $\displaystyle \widehat{TCE}=\widehat{TDE}$ (1)

    $\displaystyle CE\perp AB\Rightarrow\widehat{TCE}=\widehat{DBE}$ (2)

    $\displaystyle \Delta TCD\equiv\Delta TED\Rightarrow\widehat{TDE}=\widehat{TDC}$ (3)

    From (1), (2), (3) $\displaystyle \widehat{TDC}=\widehat{DBE}\Rightarrow TD\parallel AB\Rightarrow\widehat{TDE}=\widehat{DEB}\Rightarro w\widehat{DEB}=\widehat{DBE}$
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  3. #3
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    I would show the $\displaystyle \angle DEB \cong \angle BED$.
    This can be done by noting that both measure $\displaystyle \frac{m(\widehat{AE})}{2}$.
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  4. #4
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    Why is $\displaystyle CE\perp AB\$?

    Because, from there, I didn't understand.

    Thanks, BG
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  5. #5
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    Quote Originally Posted by BG5965 View Post
    red_dog, can you please reply, or could someone please explain my question in the above post?
    $\displaystyle m\left( {\angle EBD} \right) = \frac{{m\left( {\widehat{AC}} \right) - m\left( {\widehat{EC}} \right)}}{2}$
    $\displaystyle m\left( {\angle DEB} \right) = \frac{{m\left( {\widehat{EA}} \right)}}{2}\;\& \,m\left( {\widehat{EA}} \right) = m\left( {\widehat{AC}} \right) - m\left( {\widehat{EC}} \right)$
    But that means that $\displaystyle m\left( {\angle DEB} \right) = m\left( {\angle EBD} \right)\, \Rightarrow \,\angle DEB \cong \angle EBD$
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