Question

through the vertex A of parallelogram ABCD, a line AEF is drawn to meet BC at E and DC produced at F. Show
that the triangles BEF and DCE are equal in area.​

Answer 1

Proved below.

Step-by-step explanation:

Given:

Here, ABCD is a parallelogram and a line AEF is drawn to meet BC at E and DC produced at F.

To prove:

Δ BEF = Δ DCE

Construction: Join one diagonal AC

Proof:

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 Δ ABE  + Area of Δ 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.