Proof for Symmetry property of Congruent Segments

Question

I am starting to learn geometrical proofs, and I have come across the Symmetry property of segment congruence (if $AB$ is congruent to $CD$, then $CD$ is congruent to $AB$).

One of the exercises in my book tell me to prove this using the property of reflexivity (segment AB is congruent to segment $AB$), and the theorem that is if segment $AB$ is congruent to segment $CD$ and segment $AB$ is congruent to segment $EF$, then segment $CD$ is congruent to segment $EF$.

I am really stuck and have no idea how to prove this. Please help or give me a starting point.

Thank you.

Answer 1

Here's a more a complete answer. There are two givens, the Euclidean and reflexive properties of congruence.

i. $ \text{If}\space AB\cong CD \space\text{and}\space AB\cong EF, \space\text{then} CD\cong EF. \space\space\space\space\text{(Euclidean property)}$

ii. $ AB \cong AB.\space\space\space\space\text{(reflexive property)}$

Now, the statement you have prove is the symmetric property: $ \text{if}\space AB\cong CD, \space\text{then} \space CD \cong AB. $ To prove it, you start by assuming $AB\cong CD$. Remember you are also given $AB \cong AB$. But if both $AB\cong CD \space\text{and}\space AB\cong AB$, then by the Euclidean property $CD \cong AB$. QED.