Solve the following question using an appropriate Euclid's axiom:
In the figure:
$AB = BC$,$M$ is the mid-point of $AB$ and $N$ is the mid-point of $BC$. Show that $AM = NC$.

(N/A) Given: $AB = BC$ ... $(1)$
Since $M$ is the mid-point of $AB$,we have $AM = MB = \frac{1}{2} AB$.
Therefore,$AB = 2 AM$ ... $(2)$
Since $N$ is the mid-point of $BC$,we have $BN = NC = \frac{1}{2} BC$.
Therefore,$BC = 2 NC$ ... $(3)$
From equation $(1)$,we have $AB = BC$.
Substituting the values from $(2)$ and $(3)$ into $(1)$,we get:
$2 AM = 2 NC$
According to Euclid's axiom: "Things which are halves of the same things are equal to one another."
Since $AM = \frac{1}{2} AB$ and $NC = \frac{1}{2} BC$,and $AB = BC$,it follows that $AM = NC$.