Given: Triangles ABC, ABD equilateral; E and F midpoints of AC, BC

Prove: Points G and H trisect AB

CPCTC
CPCTC
An equilateral triangle is equiangular
KD = 2/3 JD
Definition
Definition
JK = 1/2 CK
The midsegment theorem
∠EAB = ∠FBA; ∠BAD = ∠ABD
EF ǁ AB; EF = 1/2 AB
Algebra
JK = 1/2 KD
Algebra
△AGD ≅ △BHD
AG + BH = 2/3 AB
Algebra
Algebra
The diagonals of a parallelogram bisect each other
AG = GH = HB = 1/3 AB
Triangles ABC, ABD equilateral; E and F midpoints of AC, BC
Side-Splitter theorem
Side-Splitter theorem
∠EDA = ∠FDB
∠EAD = ∠FBD
Algebra
ASA (Angle-Side-Angle)
AC = BC = BD = AD = AB
Definition
Substitution
SAS (Side-Angle-Side)
∠EAD = ∠EAB + ∠BAD; ∠FBD = ∠FBA + ∠ABD
GH = 2/3 EF
All rhombuses are parallelograms
ABCD is a rhombus
Points G and H trisect AB
Angle addition theorem
Given
ABCD is a parallelogram
AK = KB; CK = KD
GH = 1/3 AB
△EAD ≅ △FBD
AG = BH