Question

In figure $ABCD$, $ABFE$ and $CDEF$ are parallelograms. Prove that $ar\left(△ADE\right)=ar(△BCF)$

Answer 1

In $\Delta ADE$ & $\Delta BCF\to$

$AD=BC(ABCD$ is a parallelogram) (opp. sides of $11gm)$

$AE=BE(ABFE$ is a parallelogram) (opp. sides of $11gm)$

$ED=FC(CDEF$ is a $11gm)$ (opp. sides of $11gm)$

$\therefore \Delta ADE\cong \Delta BCF$ (by SSS)

Since $\Delta ADE$ & $\Delta BCF$ are congruent

$arc\left(\Delta ADE\right)=arc\left(\Delta BCF\right)$