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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