#
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.

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 $\u2206$ ABC = Area of $\u2206$ 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 $\u2206$ ABE + Area of $\u2206$ AEC = Area of $\u2206$ ABE + Area of $\u2206$ BEF , So

Area of $\u2206$ AEC = Area of $\u2206$ BEF ------------- ( 1 )

And

Area of $\u2206$ AEC = Area of $\u2206$ 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 $\u2206$ BEF = Area of $\u2206$ DCE ( Hence proved )**

**
**