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(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

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(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 $AM = NC$.

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 $AM = NC$.

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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 $AM = NC$.