See complete Problem 50

Triangle with Equilateral triangles. Level: High School, SAT Prep, College geometry

Post your solutions or ideas in the comments.

## Monday, May 19, 2008

### Elearn Geometry Problem 50

Labels:
30-60,
angle,
angle bisector,
equilateral,
perpendicular,
triangle

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1) ▲DBC = ▲ABE ( BC=BE, AB=BD, DBC=EBA=60+B ) =>

ReplyDeleteAE = 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