In the given figure, $\mathrm{AD}\parallel \mathrm{BE}\parallel \mathrm{CF}$. Then choose the correct one.

Here, $∆\mathrm{BAE}$ and $∆\mathrm{BDE}$ are on the same base BE and between the same parallel BE  and AD.

Since two triangles on the same base and between the same parallel are equal in area,

Area of  $∆\mathrm{BA}\text{E}$ = Area of $(∆\mathrm{BDE})$      ......(1) 

$\text{∆BCE}$ and $∆\mathrm{BEF}$ are on the same base BE and between the same parallel BE and CF.

Since two triangles on the same base and between the same parallel are equal in area,

Area of  $\text{∆BCE}$ = Area of $∆\mathrm{BEF}$      ......(2) 

Adding the corresponding sides (1) and (2) we get, 

Area of  $∆\mathrm{BA}\text{E}$ + Area of  $\text{∆BCE}$ = Area of $(∆\mathrm{BDE})$  + Area of $∆\mathrm{BEF}$  ....(3)

From figure,

Area of $∆\mathrm{AEC}$ = Area of  $∆\mathrm{BA}\text{E}$ + Area of  $\text{∆BCE}$ 

Also,

Area of $∆\mathrm{DBF}$ = Area of $(∆\mathrm{BDE})$  + Area of $∆\mathrm{BEF}$ 

Therefore, in equation (3), we get:

Hence, $∆\mathrm{AEC}=∆\mathrm{DBF}$.