5 thoughts on “Similar Triangles”

  1. Let ∠ACB be a;
    Since ∠EAF = 90, then ∠CAB= 90;
    ∠CED = 180-90-a
    = 90-a (angle sum of triangle = 180)
    ∠CBA = 180-90-a
    = 90-a (angle sum of triangle = 180)
    ∠FEA = 90-a (vertically opposite)
    ∠EFA = 180-(90-a)
    = 90+a (angle sum of triangle=180)
    ∠AED = 180-(90-a)
    = 90+a (supplementary)

    Proof for similarity between Δ DEA ||| Δ ADF;
    A = ∠EFA = ∠DEA
    A = ∠EDA = ∠EDA

    Thus, Δ DEA ||| Δ ADF

        1. lol. yup that’s what i mean. hahah. i miscalculate it.
          Point D seems to be the midpoint of line CB. Thus:
          ∠EAD = a,
          therefore ∠EFA = ∠EAD

          Then, we can get:
          ∠FAD = 90+a;
          ∠AED = 180-(90-a)
          = 90+a (supplementary);
          ∠FAD =∠AED

          Proof for similarity between Δ DEA ||| Δ ADF;
          A = ∠FAD =∠AED
          A = ∠EDA = ∠EDA

          Thus, Δ DEA ||| Δ ADF 🙂

          uggh bener gaakk sih?

          1. Thanks ya both Joshua and Jonathan. Both are correct 🙂
            Check on my latest question / quiz after this 🙂

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