Question 13 Marks
In the we have $AC = DC, CB = CE$. Show that $AB = DE.$


Answer
View full question & answer→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 Eqs.$(i)$ and $(ii)$, we get $AC + CB = DC + CE \Rightarrow AB = DE$

