Medium
In the figure,if $AC = BD$,then prove that $AB = CD$.
(N/A) We have,
$AC = BD$ [Given] .......... $(1)$
Since the point $B$ lies between $A$ and $C$,
$\therefore AC = AB + BC$ .......... $(2)$
Similarly,since the point $C$ lies between $B$ and $D$,
$\therefore BD = BC + CD$ .......... $(3)$
From $(1)$,$(2)$,and $(3)$,
$AB + BC = BC + CD$
Subtracting $BC$ from both sides (Euclid's axiom: If equals are subtracted from equals,the remainders are equal),
$\Rightarrow AB = CD$.
$AC = BD$ [Given] .......... $(1)$
Since the point $B$ lies between $A$ and $C$,
$\therefore AC = AB + BC$ .......... $(2)$
Similarly,since the point $C$ lies between $B$ and $D$,
$\therefore BD = BC + CD$ .......... $(3)$
From $(1)$,$(2)$,and $(3)$,
$AB + BC = BC + CD$
Subtracting $BC$ from both sides (Euclid's axiom: If equals are subtracted from equals,the remainders are equal),
$\Rightarrow AB = CD$.
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