## 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.

Complete Problem 526

Level: High School, SAT Prep, College geometry

1. Area(BDE)+Area(EDC)=Area(BDC)
½*e.d.sin(60)+1/2*d.f.sin(60)=1/2*e.f*sin(120)
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
Ajit

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.

Migue.

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)
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
Moratuwa
Sri Lanka

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