Saturday, October 9, 2010

Problem 526: Equilateral Triangle, Chord, Measurement

Proposed Problem
Click the figure below to see the complete problem 526 about Equilateral Triangle, Chord, Measurement.

 Problem 526: Equilateral Triangle, Chord, Measurement.
See also:
Complete Problem 526

Level: High School, SAT Prep, College geometry


  1. Area(BDE)+Area(EDC)=Area(BDC)
    So e.d+d.f=e.f
    Divide both side by e.d.f we get the result

    Peter Tran

  2. If we must eschew trigonometry then, by Ptolemy's Theorem, we've: AB*f+ AC*e=AD*BC or AD = e + f----(1). Further,/_BED=/_ABC+/_BAD=/_ACB +/_BCD=/_ACD and /_EBD=/_CAD or Tr.CAD /// Tr.BED ----> f/d=AD/e or 1/d=AD/(e*f) ---(2). Eqns. (1) & (2) give us the necessary result viz. 1/d = 1/e + 1/f

  3. Let BE = m and EC = n. Let a side of ABC
    Tr. ACE similar to Tr. BDE -> a/e = n/d
    Tr. AEB similar to Tr. CDE -> a/f = m/d
    Then, a/e + a/f = (m+n)/d -> 1/e + 1/f = 1/d.


  4. since ABCD is a cyclic quadrilateral, by ptolemy's theorem, we have AD*BC = AB*f + AC*e ... (i)
    since AB = BC = CA as triangle ABC is equilateral,
    we have AD = e + f ... (ii)
    triangle BDE ~ triangle ADC.
    which implies that BD/AD = DE/DC or DE = DC*BD/AD
    using symbols, f for DC, d for DE and e for BD, and (ii), we get d = f*e/(f+e) or 1/d = (f+e)/f*e
    or 1/d = 1/f + 1/e
    Q. E. D.

  5. ABC = ADC = 60
    BDA = ACB = 60
    S(BDC) = S(BDE) + S(DCE)
    e.f.1/2.sen120 = e.d.1/2.sen60 + d.f.sen60.1/2
    e.f.1/2.sen120 = d.sen60.1/2(e+f)
    ef = d(e+f)
    d = ef/e+f
    d = 1/e +1/f

  6. Draw BF // to AD where F is on CD extended. Tr. BFD is easily seen to be equilateral and from similar Tr.s e/d = (e+f)/f. Divide both sides of the equation by def and the result follows.

    Sumith Peiris
    Sri Lanka

  7. Applying Prolemy to cyclic quad ABDC we can also that AD = e + f