# Q.3. (b) in the figure given above, DE ‖ BC. Prove that: (i) area of ∆ ACD = area of ∆ ABE (ii) area of ∆​ OBD = area of ∆ OCE

Dear Student,

We know :  Area of triangle  = $\frac{1}{2}$ $×$ Base $×$ Height

Let , Height between parallel lines ( DE | | BC ) =  h

So,

Area of $∆$ BDE = $\frac{1}{2}$ $×$ DE $×$ h                             --- ( 1 )              ( Here base =  DE and Height = h as that is lies between DE | | BC )

And

Area of $∆$ CDE = $\frac{1}{2}$ $×$ DE $×$ h                             --- ( 2 )              ( Here base =  DE and Height = h as that is lies between DE | | BC )

From equation 1 and 2 we get :

Area of $∆$ CDE  = Area of $∆$ BDE                                          --- ( 3 )

i ) Now we add " Area of $∆$ ADE " on both hand side of equation 3 and get :

Area of $∆$ CDE + Area of $∆$ ADE  = Area of $∆$ BDE  + Area of $∆$ ADE

Area of $∆$ ACD  = Area of $∆$ ABE                                                  ( Hence proved )

ii ) Now we subtract " Area of $∆$ ODE " on both hand side of equation 3 and get :

Area of $∆$ CDE - Area of $∆$ ODE  = Area of $∆$ BDE  - Area of $∆$ ODE

Area of $∆$ OCE  = Area of $∆$ OBD                                                  ( Hence proved )