# [SOLVED] Geometry: Not a proof. Stuck on figure.

• May 13th 2009, 09:55 AM
ilikedmath
[SOLVED] Geometry: Not a proof. Stuck on figure.

If $\displaystyle \bigtriangleup KLT$ $\displaystyle \cong$ $\displaystyle \bigtriangleup KLM$ 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 $\displaystyle \neq$ T.

The hint our teacher gave was that $\displaystyle \angle$K $\displaystyle \cong$ $\displaystyle \angle$K. 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:

http://i40.tinypic.com/288322u.jpg

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(Headbang)
Any help is greatly appreciated! Thanks!
• May 13th 2009, 10:04 AM
Plato
The problem with your drawing is the $\displaystyle T~\&~M$ are on oppsite sides of $\displaystyle \overleftrightarrow {KL}$ contrary to the given.
• May 13th 2009, 10:13 AM
ilikedmath
corrected
Quote:

Originally Posted by Plato
The problem with your drawing is the $\displaystyle T~\&~M$ are on oppsite sides of $\displaystyle \overleftrightarrow {KL}$ contrary to the given.

ok, thanks. i changed the figure now. is that correct?(Thinking)
• May 13th 2009, 10:27 AM
masters
Quote:

Originally Posted by ilikedmath
ok, thanks. i changed the figure now. is that correct?(Thinking)

Hi ilikedmath,

If your initial statement is correct: $\displaystyle \triangle KLT \cong \triangle KLM$, then we have to say that $\displaystyle \angle T \cong \angle M$. They are corresponding angles in two congruent triangles.
• May 13th 2009, 11:05 AM
Plato
Masters, unfortunately your figure is not a model of the relation $\displaystyle \Delta KLT \cong \Delta KLM$.
In an axiomatic geometry course that notation is very precise. It means:
$\displaystyle \begin{array}{ccc} {K \leftrightarrow K} & {\overline {KL} \cong \overline {KL} } & {m\left( {\angle KLT} \right) = m\left( {\angle KLM} \right)} \\ {L \leftrightarrow L} & {\overline {LT} \cong \overline {LM} } & {m\left( {\angle LTK} \right) = m\left( {\angle LMK} \right)} \\ {T \leftrightarrow M} & {\overline {KT} \cong \overline {KM} } & {m\left( {\angle TKL} \right) = m\left( {\angle MKL} \right)} \\\end{array}$

Does your figure have those properties? In particular, is it true $\displaystyle {\overline {LT} \cong \overline {LM} }$?
• May 13th 2009, 11:37 AM
masters
Quote:

Originally Posted by Plato
Masters, unfortunately your figure is not a model of the relation $\displaystyle \Delta KLT \cong \Delta KLM$.

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.
• May 13th 2009, 12:30 PM
Plato
Given of the relation $\displaystyle \Delta KLT \cong \Delta KLM$, and $\displaystyle T~\&~M$ are on the same side of $\displaystyle \overleftrightarrow {KL}$.
If this is for an axiomatic geometry course, then it follows that $\displaystyle T=M$.
• May 14th 2009, 05:10 AM
ilikedmath
Thank you, masters and Plato, for your help!