How to prove point A belongs to line t?

Question

I'm stuck at trying to prove that any point $A$ will belong to line $t$ if and only if segments $AB=AC$, where $B$ and $C$ are symmetrical points to the line $t$ and $M$ is the midpoint of segment $BC$.

I started of with this graph

And modified it like so

From here I've managed to deduce that $\triangle{ABM}$ and $\triangle{ACM}$ are equal by using one attribute of equal triangles. But by this point I'm not sure this is even useful. What am I missing, how do you prove $A$ belongs to $t$ under these circumstances?

Answer 1

Consider any point $A$ that is equidistant from both $B$ and $C$. Draw a perpendicular from $A$ on BC and let it be $M'$. Consider the triangles $ABM'$ and $ACM'$, Note that in AM' is the common side of these two triangles, and $AC=AB$ is our assumption, as both of these triangles are right-angled ones, they must be congruent by our assumption, however in that case $BM'=CM'$, this implies that $M=M'$. This implies $A$ is on the line $t$.

Answer 2

Line t can be defined as a set of points whose distance from B and C are equal.

If we put the set T, A$\in$T if and only if A belongs to the line t.

So if we assume $AB \ne AC$, then A $\notin$T,

therefore A belongs to the line t if and only if AB=AC.