Prove : AX = (1/3)AB

Triangle DGM ~ Triangle BGA {by the AA Theorem of Similar Triangles}

DM = (1/2)DC {Def. of midpoint}

DC = AB {Prop. of Rectangles}

DM = (1/2)AB {Substitution}

DM/AB = 1/2 {Division}

GY/GX = 1/2 {Prop. of Similar Triangles}

Triangle MGY ~Triangle AGX {AA}

MY/AX = GY/GX {Proportional parts of similar triangles}

MY/AX = 1/2 {Substitution}

AX = DY {Prop. of rectangles}

MY/DY = 1/2 {Substitution}

MY/DY + 1 = 1/2 + 1 {Addition}

(MY + DY)/DY = 3/2 {Simplification}

DM/DY = 3/2 {Segment Addition}

DY/DM = 2/3 {Prop. of proportions}

DY = (2/3)DM = (2/3)(1/2)AB = (1/3)AB {Substitution,Multiplication}

AX = (1/3)AB {Substitution}