In the adjoining figure, DE || BC. Prove that

∆DEC and ​∆DEB lies on the same base and between the same parallel lines.
So, ar(​∆DEC) = ar(∆DEB)                      ...(1)

(i) On adding​ ar(∆ADE)​ in both sides of equation (1), we get:
ar(​∆DEC) + ar(∆ADE)​ = ar(∆DEB) + ar(∆ADE)​ ​                 
⇒ ar(​​∆ACD) = ar(​​∆ABE) 

(ii) On subtracting​ ar(ODE)​ from both sides of equation (1), we get:​
ar(​∆DEC) -">−- ar(∆ODE)​ = ar(∆DEB) -">−- ar(∆ODE)​ ​      ​
⇒ ar(​​∆OCE) = ar(​∆OBD)