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# Show that the relation $R$ defined in the set $A$ of all triangles as $R = \{(T_1, T_2) : T_1\,is\, similar\, to\, T_2\}$, is equivalence relation. Consider three right angle triangles $(T_1 \;$ with sides $3, 4, 5, \; T_2$ with sides $5, 12, 13$, and $T_3$, with sides $6, 8, 10$) Which triangles among $T_1\, T_2\, and\, T_3$ are related?

Toolbox:
• A relation R in a set A is a equivalance relation if it is symmetric, reflexive and transitive.
• A relation R in a set A is called $\mathbf{ reflexive},$ if $(a,a) \in R\;$ for every $\; a\in\;A$
• A relation R in a set A is called $\mathbf{symmetric}$, if $(a_1,a_2) \in R\;\Rightarrow\; (a_2,a_1)\in R \; for \;a_1,a_2 \in A$
• A relation R in a set A is called $\mathbf{transitive},$ if $(a_1,a_2) \in R$ and $(a_2,a_3) \in R \; \Rightarrow \;(a_1,a_3)\in R$ for all$\; a_1,a_2,a_3 \in A$
• If a triangle $T_1$ is similar to $T_2$ then $T_2$ will be similar to $T_1$
Given $R= \{ (T_1,T_2):T_1$ is similar to $T_2 \}$:
For any triangle $T_1, (T_1,T_1)\in R$ as every triangle is similar to itself. Hence $R$ is reflexive.
If any triangle $T_1$ is similar to $T_2$ then $T_2$ will be similar to $T_1$. Hence it follows that $(T_1,T_2) \in R \Rightarrow (T_2,T_1) \in R$. Therefore, $R$ is symmetric.
If $T_1$ is similar to $T_2$ and $T_2$ is similar to $T_3$ then $T_1$ is similar to $T_3$ also .
Therefore, $(T_1,T_2), (T_2,T_3) \in R \Rightarrow (T_1,T_3) \in R$. Hence, $R$ is transitive.
Since $R$ is reflexive, symmetric and transitive, it is an equivalance relation
Given three right angle triangles $T_1$ with sides $3, 4, 5, T_2$ with sides $5, 12, 13$, and $T_3$, with sides $6, 8, 10$.
We can see that the circles of triangle $T_1$ and $T_3$ are proportional because: $\frac{3}{6}=\frac{4}{8}=\frac{5}{10}=(\frac{1}{2})$
Therefore $T_1$ and $T_3$ are similar and related.
However, the circles of any two other triangles for e.g. $T_1$ and $T_2$ are not propotional as evidenced by: $\frac{3}{5} \neq \frac{4}{12} \neq \frac{5}{13}$
edited Mar 8, 2013