Question

Points E and F lie on diagonal AC of a parallelogram ABCD such that AE = CF. What type of quadrilateral is BFDE?

Answer 1

\[\text{ In the } {II}^{gm} ABCD: \]
\[AO = OC . . . . . . (i) (\text{ diagonals of a parallelogram bisect each other })\]
\[AE = CF . . . . . . . (ii) (\text{ given })\]
\[\text{ Subtracting (ii) from } (i): \]
\[AO - AE = OC - CF\]
\[EO = OF . . . . . . . . . (iii)\]
\[\text{ In } ∆ DOE \text{ and } ∆ BOF: \]
\[EO = OF (\text{ proved above })\]
\[DO = OB (\text{ diagonals of a parallelogram bisect each other })\]
\[\angle DOE = \angle BOF (\text{ vertically opposite angles })\]
\[\text{ By SAS congruence }: \]
\[ ∆ DOE \cong ∆ BOF\]
\[ \therefore DE = BF (c . p . c . t)\]
\[\text{ In } ∆ BOE \text{ and } ∆ DOF: \]
\[EO = OF (\text{ proved above })\]
\[DO = OB (\text{ diagonals of a parallelogram bisect each other })\]
\[\angle DOF = \angle BOE (\text{ vertically opposite angles })\]
\[\text{ By SAS congruence }: \]
\[ ∆ DOE \cong ∆ BOF\]
\[ \therefore DF = BE (c . p . c . t)\]
\[\text{ Hence, the pair of opposite sides are equal . Thus, DEBF is a parallelogram }. \]