Medium
Solve the following question using an appropriate Euclid's axiom:
In the figure,we have $AC = DC$ and $CB = CE$. Show that $AB = DE$.
In the figure,we have $AC = DC$ and $CB = CE$. Show that $AB = DE$.
(N/A) We have $AC = DC$ $\dots(1)$ [Given]
And $CB = CE$ $\dots(2)$ [Given]
Now,by Euclid's second axiom,if equals are added to equals,the wholes are equal.
Adding $(1)$ and $(2)$,we get:
$AC + CB = DC + CE$
Since $AC + CB = AB$ and $DC + CE = DE$,we have:
$AB = DE$.
And $CB = CE$ $\dots(2)$ [Given]
Now,by Euclid's second axiom,if equals are added to equals,the wholes are equal.
Adding $(1)$ and $(2)$,we get:
$AC + CB = DC + CE$
Since $AC + CB = AB$ and $DC + CE = DE$,we have:
$AB = DE$.
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