Medium
Solve the following question using an appropriate Euclid's axiom:
In the figure,$X$ and $Y$ are the mid-points of $AC$ and $BC$ respectively,and $AX = CY$. Show that $AC = BC$.
In the figure,$X$ and $Y$ are the mid-points of $AC$ and $BC$ respectively,and $AX = CY$. Show that $AC = BC$.
(N/A) Given: $AX = CY$ and $X, Y$ are the mid-points of $AC$ and $BC$ respectively.
Since $X$ is the mid-point of $AC$,we have $AC = 2AX$.
Since $Y$ is the mid-point of $BC$,we have $BC = 2CY$.
According to Euclid's axiom $6$,"Things which are double of the same things are equal to one another."
Since $AX = CY$,their doubles must also be equal.
Therefore,$2AX = 2CY$.
Substituting the values,we get $AC = BC$.
Since $X$ is the mid-point of $AC$,we have $AC = 2AX$.
Since $Y$ is the mid-point of $BC$,we have $BC = 2CY$.
According to Euclid's axiom $6$,"Things which are double of the same things are equal to one another."
Since $AX = CY$,their doubles must also be equal.
Therefore,$2AX = 2CY$.
Substituting the values,we get $AC = BC$.
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