Question

E and F are points on diagonal AC of a parallelogram ABCD such that AE = CF. Show that BFDE is a parallelogram.

Answer 1

Consider $\Delta AEB\&\Delta CFD$.
AE = CF
CD = AB
$\angle DCF=\angle EAB$
$\Rightarrow \Delta AEB\equiv \Delta DCF\left(congruent\right)$
$\Rightarrow EB=DF\cdots \left(1\right)\&\angle CDF=\angle ABE---\left(1a\right)$
Similary $\Delta AED\&\Delta CFB$ are congruent.
$\therefore AD=BC;AE=CF;\angle DAE=\angle FCB$
$\Rightarrow DE=FB-\left(2\right).\&\angle ADE=\angle CBF-\left(2a\right)$
$Now\angle ADC=\angle CBA(\because ABCDisparallelogram)$
$\angle ADE+\angle EDF+\angle FDC=\angle CBF+\angle FBE+\angle EBA$
$\Rightarrow \angle EDF=\angle FBE-\left(3\right)$
Similarly it can be proved that
$\Rightarrow \angle DEB=\angle DFB-\left(4\right)$
From (1), (2), (3) & (4) $\Rightarrow$ BDEF is a parallelogram