A simple example:

documentclass[12pt,border=5pt]{standalone}

usepackage{newcent,pstricks,pst-eucl} 

usepackage{auto-pst-pdf}

begin{document}

begin{pspicture}

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=60]{A}{B}{C}{O}

end{pspicture}

end{document}

The result of compiling:

Question 1:

How to get a point M such as MA=2/3AB, or more generally MA=(a/b)AB and a point M' belong to small arc CA such as arc M'A=2/3AB, or more generally M'A=(a/b)AB.

Question 2:

How to get bisector of angle A BUT are two bisectors or three bisectors.

Responding to AS'comment below.

Step 1

documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 



begin{document}

begin{pspicture}[showgrid](-1,-3)(7,5)

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=60]{A}{B}{C}{O}

pstHomO[HomCoef=0.6666]{A}{B}[M]% 2/3 --> 0.6666

end{pspicture}

end{document}

Note: HomCoef cannot accept RPN 2 3 div so I have to insert 0.6666 (3 decimals places should suffice, I think).

Step 2

documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 



begin{document}

begin{pspicture}[showgrid](-1,-3)(7,5)

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=60]{A}{B}{C}{O}

pstHomO[HomCoef=0.6666]{A}{B}[M]% 2/3 --> 0.6666

pstRotation[AngleCoef=0.3333,RotAngle=pstAngleAOB{C}{O}{A}]{O}{C}[M']

end{pspicture}

end{document}

Note: AngleCoef must come before RotAngle. It is not commutative!

Step 3 (Final)

documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 



begin{document}

begin{pspicture}[showgrid=false](-1,-3)(7,5)

pstTriangle(2,4){A}(0,0){B}(6,0){C}

pstCircleABC[PosAngle=225]{A}{B}{C}{O}

pstHomO[HomCoef=0.6666]{A}{B}[M]% 2/3 --> 0.6666

pstRotation[AngleCoef=0.3333,RotAngle=pstAngleAOB{C}{O}{A}]{O}{C}[M']

psset{PointName=none,PointSymbol=none}

pstRotation[AngleCoef=0.3333,RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[P1]

pstRotation[AngleCoef=0.6666,RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[P2]

pstInterLL{B}{C}{A}{P1}{Q1}

pstInterLL{B}{C}{A}{P2}{Q2}

psset{Mark=MarkHash,MarkAngle=90,MarkAngleRadius=.8}

pstMarkAngle{B}{A}{Q1}{}

pstMarkAngle{Q1}{A}{Q2}{}

pstMarkAngle{Q2}{A}{C}{}

psset{linestyle=dashed}

psline(A)(Q1)

psline(A)(Q2)

end{pspicture}

end{document}

Note: We have pstSegmentMark (ends with Mark) to mark a segment but we have pstMarkAngle (begins with Mark) to mark an angle. It seems the package author likes making inconsistent names.

documentclass[pstricks,12pt,border=15pt]{standalone}

usepackage{pst-eucl} 



begin{document}

begin{pspicture}[showgrid](-1,-3)(7,5)

  pstTriangle(2,4){A}(0,0){B}(6,0){C}

  pstCircleABC[PosAngle=60]{A}{B}{C}{O}

  pstHomO[HomCoef=0.667]{A}{B}[M]

  pstRotation[AngleCoef=0.333,RotAngle=pstAngleAOB{C}{O}{A}]{O}{C}[M']

  psset{PointSymbol=none,PointName=none}

  pstRotation[AngleCoef=0.333,RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[M1]

  pstRotation[AngleCoef=0.667,RotAngle=pstAngleAOB{B}{A}{C}]{A}{B}[M2]

  pcline[linestyle=dashed](A)(M1)

  pcline[linestyle=dashed](A)(M2)

end{pspicture}

end{document}