Q2 . Through vertex of A of a llgm drawn to meet BC at E and DC produced at F. Show that triangle BEF and triangle CDE are equal in area.

Answer :

From given information we draw our diagram , As  :

We also join one diagonal AC ,

We know from parallelogram property 
AB  =  CD  , AB  | |  CD
And
BC  =  DA ,  BC | |  DA
We know Area of triangle  =  $\frac{1}{2}\times Base \times Height$
So ,

Area of $∆$ ABC  =  Area of $∆$ ABF  ( As both have same base ( AB )  and same height as they lies between two parallel lines AB and CD )

Now we can write these areas , As  :

Area of $∆$ ABE  +  Area of $∆$ AEC  =  Area of $∆$ ABE  + Area of $∆$ BEF   , So

Area of $∆$ AEC  =  Area of $∆$ BEF                                                   ------------- (  1  )

And

Area of $∆$ AEC  =  Area of $∆$ DCE                                                   ------------- (  2  )     ( As both have same base ( CE )  and same height as they lies between two parallel lines BC and DA )

From equation 1 and 2 , we get 

Area of $∆$ BEF =  Area of $∆$ DCE                                                      ( Hence proved )