Points E and F lie on diagonal AC of a parallelogram ABCD such that AE = CF.

Solution:

In the IIgm $\mathrm{ABCD}:$

$\mathrm{AO}=\mathrm{OC} \ldots \ldots$ (i) (diagonals of a parallelogram bisect each other)

$\mathrm{AE}=\mathrm{CF} \ldots \ldots$ (ii) (given)

$\mathrm{Subtracting}$ (ii) from (i):

$\mathrm{AO}-\mathrm{AE}=\mathrm{OC}-\mathrm{CF}$

$\mathrm{EO}=\mathrm{OF} \ldots \ldots$ (iii)

$\mathrm{In} \Delta \mathrm{DOE}$ and $\Delta \mathrm{BOF}:$

$\mathrm{EO}=\mathrm{OF}$ (proved above)

$\mathrm{DO}=\mathrm{OB}$ (diagonals of a parallelogram bisect each other)

$\angle \mathrm{DOE}=\angle \mathrm{BOF}$ (vertically opposite angles)

By SAS congruence:

$\Delta \mathrm{DOE} \cong \Delta \mathrm{BOF}$

$\therefore \mathrm{DE}=\mathrm{BF}$ (c. p.c.t)

In $\Delta \mathrm{BOE}$ and $\Delta \mathrm{DOF}:$

$\mathrm{EO}=\mathrm{OF}$ (proved above)

$\mathrm{DO}=\mathrm{OB}$ (diagonals of a parallelogram bisect each other)

$\angle \mathrm{DOF}=\angle \mathrm{BOE}$ (vertically opposite angles)

By SAS congruence :

$\Delta \mathrm{DOE} \cong \Delta \mathrm{BOF}$

$\therefore \mathrm{DF}=\mathrm{BE}$ (c. p.c.t)

Hence, the pair of opposite sides are equal. Thus, DEBF is a parallelogram.