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Math Help - [SOLVED] Geometry: Not a proof. Stuck on figure.

  1. #1
    Member ilikedmath's Avatar
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    Question [SOLVED] Geometry: Not a proof. Stuck on figure.

    The problem reads:

    If \bigtriangleup KLT \cong \bigtriangleup KLM and M and T are on the same side of line KL, must M = T?
    If so, why? If not, sketch a figure in which M \neq T.

    The hint our teacher gave was that \angleK \cong \angleK. So I attempted to draw a figure to help me figure this out, but I got stuck with the criterion that M and T are on the same side of line KL.

    Here's my figure:




    Should I have drawn two separate triangles?
    I just am quite confused with what the instructions are specifically asking.
    Our teacher said this problem should not take much time since it is not a proof, and if we've been working on it for more than 5 minutes then we're overthinking it.

    I've been stuck on this problem since yesterday afternoon
    Any help is greatly appreciated! Thanks!
    Last edited by ilikedmath; May 13th 2009 at 11:13 AM. Reason: correction after first response
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  2. #2
    MHF Contributor

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    The problem with your drawing is the T~\&~M are on oppsite sides of \overleftrightarrow {KL} contrary to the given.
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  3. #3
    Member ilikedmath's Avatar
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    corrected

    Quote Originally Posted by Plato View Post
    The problem with your drawing is the T~\&~M are on oppsite sides of \overleftrightarrow {KL} contrary to the given.
    ok, thanks. i changed the figure now. is that correct?
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  4. #4
    A riddle wrapped in an enigma
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    Quote Originally Posted by ilikedmath View Post
    ok, thanks. i changed the figure now. is that correct?
    Hi ilikedmath,

    If your initial statement is correct: \triangle KLT \cong \triangle KLM, then we have to say that \angle T \cong \angle M. They are corresponding angles in two congruent triangles.
    Last edited by masters; May 13th 2009 at 12:36 PM. Reason: Thanks to Plato for pointing out the incorrect drawing
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  5. #5
    MHF Contributor

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    Masters, unfortunately your figure is not a model of the relation \Delta KLT \cong \Delta KLM.
    In an axiomatic geometry course that notation is very precise. It means:
    \begin{array}{ccc}<br />
   {K \leftrightarrow K} & {\overline {KL}  \cong \overline {KL} } & {m\left( {\angle KLT} \right) = m\left( {\angle KLM} \right)}  \\<br />
   {L \leftrightarrow L} & {\overline {LT}  \cong \overline {LM} } & {m\left( {\angle LTK} \right) = m\left( {\angle LMK} \right)}  \\<br />
   {T \leftrightarrow M} & {\overline {KT}  \cong \overline {KM} } & {m\left( {\angle TKL} \right) = m\left( {\angle MKL} \right)}  \\\end{array}

    Does your figure have those properties? In particular, is it true {\overline {LT}  \cong \overline {LM} }?
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    A riddle wrapped in an enigma
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    Quote Originally Posted by Plato View Post
    Masters, unfortunately your figure is not a model of the relation \Delta KLT \cong \Delta KLM.
    You are right, Plato. I'm not even sure it can be drawn according to the given specs. I shall dispose of the offending attachment post haste. Thanks.
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  7. #7
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    Given of the relation \Delta KLT \cong \Delta KLM, and T~\&~M are on the same side of \overleftrightarrow {KL}.
    If this is for an axiomatic geometry course, then it follows that T=M.
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  8. #8
    Member ilikedmath's Avatar
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    Thank you, masters and Plato, for your help!
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