Medium
Solve the following question using appropriate Euclid's axiom:
In the figure:
$BM = BN$,$M$ is the mid-point of $AB$ and $N$ is the mid-point of $BC$. Show that $AB = BC$.
In the figure:
$BM = BN$,$M$ is the mid-point of $AB$ and $N$ is the mid-point of $BC$. Show that $AB = BC$.
(N/A) Given: $BM = BN$ ... $(1)$
Since $M$ is the mid-point of $AB$,we have $AM = BM$ ... $(2)$
Since $N$ is the mid-point of $BC$,we have $BN = NC$ ... $(3)$
From $(1)$,$(2)$,and $(3)$,we can say that $AM = NC$ ... $(4)$
Now,adding $(1)$ and $(4)$,we get:
$BM + AM = BN + NC$
Since $BM + AM = AB$ and $BN + NC = BC$,we have:
$AB = BC$
This is justified by Euclid's second axiom: "If equals are added to equals,the wholes are equal."
Since $M$ is the mid-point of $AB$,we have $AM = BM$ ... $(2)$
Since $N$ is the mid-point of $BC$,we have $BN = NC$ ... $(3)$
From $(1)$,$(2)$,and $(3)$,we can say that $AM = NC$ ... $(4)$
Now,adding $(1)$ and $(4)$,we get:
$BM + AM = BN + NC$
Since $BM + AM = AB$ and $BN + NC = BC$,we have:
$AB = BC$
This is justified by Euclid's second axiom: "If equals are added to equals,the wholes are equal."
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