In the given figure, $AD\parallel BE\parallel CF$.

Concept: It is known that if two triangles have the same base and lie between the same parallels, their area is equal.

Here, $∆BAE$ and  $∆BDE$ 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 areas,

Area of  $∆BAQ=Area of (∆BDE)......\left(1\right)$ 

$∆BCE$ and $∆BEF$ 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 areas,,

$\therefore Area of (∆BCE)= Area of ∆BEF$ …..(ii) 

When adding the corresponding sides (i) and (ii) we get, 

$Area of ∆BAQ+Area of ∆BCE$ Area of 

 $= Area of ∆BDE+ Area of ∆BEF$ 

Hence, 

$∆AEC=∆DBF$.