Prove that if the identity is written as the product of $r$ transpositions, then $r$ is an even number

Notice that in each of the three given identities

$(bc)(ab)=(ac)(bc)$

$(cd)(ab)=(ab)(cd)$

$(ac)(ab)=(ab)(bc)$

if you replace the left side with the right side, there is no longer an $a$ in the second parentheses. This means that each time you apply any of these identities, the rightmost occurrence of $a$ is one transposition further to the left.

Eventually either you will get two identical adjacent transpositions $(ax)(ax)=id$, so they disappear and you can apply induction to the shorter string of transpositions; or you end up with $a$ only appearing in the leftmost transposition of the entire product of transpositions, but this is impossible since then the permutation doesn't fix $a$, and so is not the identity.

You can identify a permutation with a permutation matrix, i.e., the one you get by permuting the columns of the identity matrix according to the permutation. Composition of permutations corresponds to matrix multiplication, and since the determinant of a transposition is $-1$, the determinant of a composition of $r$ transpositions is $(-1)^r$. So if it is the identity, then $(-1)^r=1$, implying that $r$ is even.

You could try using the signature of the permutations. If $id=\prod_{i=1}^r \tau_i$ where $\tau_i$ are transpositions, you have that $\varepsilon(id)=1$ and $\varepsilon(\tau_i)=(-1)$, therefore $(-1)^r=1\Rightarrow r$ even.

I do not know if you studied signatures.