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

Prove: Points G and H trisect AB

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