## Tuesday, February 24, 2009

### Problem 256: Equilateral Triangle, Circumcircle, Point, Vertices, Distances

Proposed Problem

See complete Problem 256 at:
gogeometry.com/problem/p256_equilateral_triangle_circumcircle_distance_vertex.htm

Level: High School, SAT Prep, College geometry

1. Use Ptolemy's therem by which in this case we've:
AD*BC = CA*BD + AB*CD or d*BC = e*CA + f*AB which leads us to d = e + f since AB = BC = CA
Ajit: ajitathle@gmail.com

2. according to ptolemy's theorem, AB.CD + BD.AC = BC.AD. but triangle ABC is equilateral. so AB = BC = CA so we have AD = BD + CD that is d = e + f
Q. E. D.

3. Obviously ∠BDA = ∠ADC = 60°.

Let B→B'. Then ΔBDB' is equilateral, and DB' = e.

Then B'→D, A→C, so B'A = DC = f.

As a result, d = DA = DB' + B'A = e + f.

4. Problem 256 (solution 1)
Following on the DB to point B take the point E such that DE=AD (<EDA=<BDA=<BCA=60).Then triangleADE is equilateral .But triangleAEb=triangleADC
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

5. Problem 256 (solution 2)
Following on the BD to point D take the point F such that DF=DC (<BDA=<CDA=<CDF=60).Then triangleCDF is equilateral .But triangleBCF=triangleADC
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

6. Problem 256 (solution 3)
On the side AB ( not to extend ) get the point K such that BD=BK.Then triangle BKD is equilateral(<BDK=60=
<BKD ),so triangleABK=triangleCBD (AB=BC,BK=BD,<ABK=60-KBC=<CBD), so AK=DC. Therefore AD=AK+KD=DC+DB.
APOSTOLIS MANOLOUDIS 4 HIGH SHCOOL OF KORYDALLOS PIRAEUS GREECE

7. Extend CD to F such that BF//AD.
Then Tr. BDF is equilateral and
Tr.s ABD ≡ CBF.

Hence AD = CF = CD + DF so d = e + f

Sumith Peiris
Moratuwa
Sri Lanka

8. See the drawing

∠ACD= ∠ACB+ ∠BCD=Π/3+ ∠BCD
And ∠ACD= ∠ACD’+ ∠D’CD=∠BCD+ ∠D’CD => ∠D’CD = Π/3
CDA intercepts arc CA such as ABC => ∠CDA= ∠ABC=Π/3
∠D’CD = Π/3 and ∠CDA= ∠CDD’= Π/3 => ΔD’CD is equilateral
=>D’D=CD=e

9. Draw AE such that AE=e and forms a regular trapezium ADBE
BDC and AEB are congruent => EB=f
Drop Perpendiculars from B and E on AD and denote them as B' and E'
=>B'E'=f
Since m(BCA)=60=> m(BDA)=60=>Tr.BDB' is a 30-60-90 triangle
=>DB'=E'A=e/2

10. See diagram here.
∆ABE ~ ∆CDE ~ ∆ADB and ∆AEC ~ ∆BED
Therefore a/AE = f/CE = d/a and a/AE = e/BE
So a/AE = f/CE = e/BE = (f+e)/(BE+EC) = (f+e)/a = d/ad = e+f

11. Here is my solution...