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ABCD is s rhombus. EABF is …

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ABCD is s rhombus. EABF is a straight line such that EA=AB=BF. Prove that ED and FC ,when produced ,meet at right angles.

    • Posted by Ramesh Tiwari 4 years, 11 months ago

    CBSE > Class 09 > Mathematics
    • 1 answers

    Gaurav Seth 4 years, 11 months ago

    SinceΔEAD=ΔABC
    (Corresponding angles),&AD=BC(sides of the rhombus)
    By side angle side congrugency,the triangles
    Therefore the corresponding parts of the triangle are also equal.
    ΔCBF & ΔDAB
    Since the sides 
    BF=AB
    ∠CBF∠DAB(corresponding angles)
    ΔCBF&ΔDAB are congrugent
    ΔAOB
    ∠ABO+∠BAO+∠AOB=180°
    ∠ABO+∠BAO=90°
    ∠CFB+∠AED=90°
    ∠XFE+∠XEF=90°

    Read more on Brainly.in - https://brainly.in/question/987606#readmore

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