Question

ABCD is a parallelogram, AD is produced to E so that DE = DC and EC produced meets AB produced in F. Prove that BF = BC.

Answer 1

Draw a parallelogram ABCD with AC and BD intersecting at O

Produce AD to E such that DE = DC

Join EC and produce it to meet AB produced at F.

In ΔDCE ,

∴`∠`DCE = `∠`DEC    .........CD     [In a triangle, equal sides have equal angles opposite]

AB || CD                                         (Opposite sides of the parallelogram are parallel)

∴AE || CD    ( AB Lies on  AF )

AF || CD and EF is the transversal.

∴`∠`DCE = `∠`BFC    .....(2)         [Pair of corresponding angles]

From (1) and (2), we get

`∠`DEC = `∠`BFC

In ΔAFE,

`∠`AFE = `∠`AEF         (`∠`DEC = `∠`BFC )

∴AE = AF    (In a triangle, equal angles have equal sides opposite to them)       

⇒  AD + DE = AB + BF

⇒  BC + AB = AB + BF                 [ ∵ AD = BC, DE = CD and CD = AB, AB = DE ]

⇒   BC = BF