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

Prove: Points G and H trisect AB

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