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• 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

  • Josh, please check on the angle EFA :)

    • Angle EFA is 180 - ( 90 - a ) - 90 or a.

      • 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?

        • Thanks ya both Joshua and Jonathan. Both are correct :)
          Check on my latest question / quiz after this :)