## Monday, May 19, 2008

### Elearn Geometry Problem 50

See complete Problem 50
Triangle with Equilateral triangles. Level: High School, SAT Prep, College geometry

#### 1 comment:

1. 1) ▲DBC = ▲ABE ( BC=BE, AB=BD, DBC=EBA=60+B ) =>
AE = DC
2) BAE = BDC (from 1) => *ADBO cyclic*
=> DAO + DBO = 180 => 60+BAO+60+ABO=120+(BAO+ABO)
=120+(BDO+ABO) => BDO+ABO=60
▲DBO => 60+BDO+ABO+BOD=180 =60+60+BOD=180 => BOD=60
in the same way
BOE = 60 =>
=>
DOE = 120
3) FH = 1/2 DC, HG = 1/2 AE (as middle line)
FH = HG ( DC = AE )
4) FHG = 120 ( angle between parallel line)
5) HFG = HGF = 60 : 2 = 30 (▲FHG isoceles)
6) see 2)
7) OBC = OEC ( have same arc, EBOC cyclic)
OEC = HGC corresponding angles
in the same way ADO = ABO = AFH
8) see 2)
9) AOB = 120 ( AOB + D = 180, cyclic)
BOC = 120 in the same way
10)CMG = HGF = 30 = HFG = HGF as corresponding angles
11)OB perpendicular to FG
from ▲MOT ( T , BO meet MN ) GMC = 30, MOB=60 =>
MTO = 90
12) DOA = 60 => POQ = 60
LPN = 60 => OPQ = 60
=> OPQ equaliteral
13) OPQ = 60, JHP = 60
=> JPH equilateral
14) from 12) and 13)
15) JH = HP = HQ + QP = HK + OP
16) HL = 1/2 HG = 1/4 AE ( LGH = 30, & HG = 1/2 AE )
17) FL = 1/2 FG
▲FLH => FL² = FH² - HL²
FL² = 1/4 DC² - 1/16 DC² (LH=1/2 FH)
FL² = 3/16 DC²
FL = (DC/4)√3 (DC=AE, from 1)
FL = (AE/4)√3