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}