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Math Help - Deductive geometry, tangents

  1. #1
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    Deductive geometry, tangents

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

    Let ABC be a right-angled triangle. A circle T having side AC as its diameter meets hypotenuse AB at point E. A tangent line to T at point E meets side BC at point D. Prove that triangle 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 \widehat{TCE}=\widehat{TDE} (1)

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

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

    From (1), (2), (3) \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 \angle DEB \cong \angle BED.
    This can be done by noting that both measure \frac{m(\widehat{AE})}{2}.
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  4. #4
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    Why is 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?
    m\left( {\angle EBD} \right) = \frac{{m\left( {\widehat{AC}} \right) - m\left( {\widehat{EC}} \right)}}{2}
    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 m\left( {\angle DEB} \right) = m\left( {\angle EBD} \right)\, \Rightarrow \,\angle DEB \cong \angle EBD
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