ABCD is a square. The bisector of ∠DBC cuts AC and CD at E and F respectively. Prove that BF × CE = BE × DF

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ABCD is a square. The bisector of ∠DBC cuts AC and CD at E and F respectively. Prove that BF × CE = BE × DF

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Hailey 4 weeks 2021-08-23T06:12:53+00:00 1 Answer 0 views 0

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    2021-08-23T06:14:34+00:00

    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

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