\problemdata     
{relations-1}           
{equivalence relation, simple}              
{source: Velleman \S 4.6, Exercise 5.}              
{S02, week 3} 		
{S02, Radhika}        	


\begin{problem} Given $T$, the set of all triangles, and $S=\{(s,t)\in
T\times T$ $|$ the triangles $s$ and $t$ are similar$\}$, prove that
$S$ is an equivalence relation.  Note: triangles $A$ and $B$ are
similar if all three angles of $A$ are equal to the angles of $B$.

%% (NOTE: Maybe replace with Rosen 6.5.32)

\solution{\textbf{Proof:} To prove this, we need to show that $S$ is
reflexive, symmetric, and transitive.  Here we go:

First of all, two triangles $s$ and $t$ are said to be
\textit{similar} if they have three pairs of equal-degree angles.
$S$ is reflexive because every triangle is similar to itself.  That
is, every triangle $s$ has three angles of equivalent size to its own
three angles.

$S$ is symmetric because if $s$ is similar to $t$, then $s$ and $t$
have angles of the same size, so $t$ is similar to $s$ as well.

Finally, $S$ is transitive because if triangle $s$ is similar to
triangle $t$, and triangle $t$ is similar to triangle $r$ then
triangles $s$ and $t$ have three pairs of equal-degree angles, and
triangles $t$ and $r$ also have three pairs of equal-degree angles.
$s$ and $r$ must share these three pairs of equal-degree angles
since triangles only have three angles. $\qed$.
}
\end{problem}

