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