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

1. ## [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 $\angle$K $\cong$ $\angle$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:

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!

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

3. ## corrected

Originally Posted by Plato
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?

4. Originally Posted by ilikedmath
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.

5. 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:
$\begin{array}{ccc}
{K \leftrightarrow K} & {\overline {KL} \cong \overline {KL} } & {m\left( {\angle KLT} \right) = m\left( {\angle KLM} \right)} \\
{L \leftrightarrow L} & {\overline {LT} \cong \overline {LM} } & {m\left( {\angle LTK} \right) = m\left( {\angle LMK} \right)} \\
{T \leftrightarrow M} & {\overline {KT} \cong \overline {KM} } & {m\left( {\angle TKL} \right) = m\left( {\angle MKL} \right)} \\\end{array}$

Does your figure have those properties? In particular, is it true ${\overline {LT} \cong \overline {LM} }$?

6. Originally Posted by Plato
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.

7. 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 $T=M$.

8. Thank you, masters and Plato, for your help!