# In the Fig. 5.10, we have AC = DC, CB = CE. Show that AB = DE.

Given: AC =DC …(i)

and CB = CE …(ii)

According to Euclid’s axiom, if equals are added to equals, then wholes are also equal.

So, on adding Eq.(i) and (ii),

We get,

AC + CB = DC + CE

AB = DE

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