ABCD is a square. The bisector of ∠DBC cuts AC and CD at E and F respectively. Prove that BF × CE = BE × DF Question ABCD is a square. The bisector of ∠DBC cuts AC and CD at E and F respectively. Prove that BF × CE = BE × DF in progress 0 Math Hailey 4 weeks 2021-08-23T06:12:53+00:00 2021-08-23T06:12:53+00:00 1 Answer 0 views 0

## Answers ( )

Given:ABCD is a square. The bisector of ∠DBC cuts AC and CD at E and F respectively.

To find:Prove that BF × CE = BE × DF

Solution:In the figure, ABCD is a square.

∠A = ∠B = ∠C = ∠D = 90°

Construction: Draw diagonals AC and BD.

∴ ∠DBC = ∠BCA = ∠BDF = 45° ….. ( 1 ) (The diagonals of square bisect the opposite angles)

Construction: Draw the bisector of ∠DBC which cuts AC & CD at E and F respectively.

∴ ∠DBF = ∠CBF …..( 2 )

Now consider,

In ΔBDF and ΔBCE,

∠BDF ≅ ∠BCA [ using ( 1 ) ]

∠DBF ≅ ∠CBF [ using ( 2 ) ]

∴ ΔBDF ∼ ΔBCE [using AA postulate]

BD / BC = DF / CE = BF / BE [Corresponding Sides of Similar Triangles]

DF / CE = BF / BE

DF × BE = BF × CE

Hence it is proved that BF × CE = BE × DF