Question

In the given figure,
$AD||BE||CF.$
Prove that :
$Area(△AEC)=Area(△DBF)$

Answer 1

$△BAE$ and $△BDE$ are on the same base BE and between the same parallel BE and AD.
$\therefore$ Area of $△BAQ$=Area of $△BDE$ [two triangle on the same base and between the same parallel are equal in areas] (I)
$△BCE$ and $△BEF$ are on the same base BE and between the same parallel BE and CF.
$\therefore$ Area of $△BCE$=Area of $△BEF$ [two triangle on the same base and between the same parallel are equal in areas] (II)
Adding the corresponding sides (i) and (ii) we get,
Area of $△BAQ$+Area of $△BCE$=Area of $△BDE$+Area of $△BEF$
=>$△AEC$=$△DBF$