Adv
  1. Given that. AB II CD In Δ AOB and Δ COD ∠OAB = ∠OCD (Alternate interior angles) ∠OBA = ∠ODC (Alternate interior angles) ∠AOB = ∠DOC (Vertically opposite angles) ΔAOB ~ ΔCOD (AAA similarly) We know that, Areabof ΔAOB/Area of ΔCOD ​(AB/CD​)2=(AO/CO​)2=(OB/OD​)2 (2CD/CD​)2 = 4:1

    Given that.
    AB II CD
    In Δ AOB and Δ COD
    OAB = OCD (Alternate interior angles)
    OBA = ODC (Alternate interior angles)
    AOB = DOC (Vertically opposite angles)
    ΔAOB ~ ΔCOD (AAA similarly)
    We know that,
    Areabof ΔAOB/Area of ΔCOD (AB/CD)2=(AO/CO)2=(OB/OD)2
    (2CD/CD)2 = 4:1

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