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Just (a) and (b) for now . a). Let $Oxy$ be the angle. Align one of the sides of the ruler along $(Ox$ s.t. the other side cuts $(Oy$ in $Y$. We can use this side to draw a parallel $\ell_y$ through $Y$ to $Ox$. Do the same by interchanging the roles of $(Ox,(Oy$ to find the point $X\in (Ox$ and the parallel $\ell_x$ through $X$ to $Oy$. The line passing through $O$ and the point of intersection of $\ell_x,\ell_y$ is the bisector. b). Call the segment $AB$. We can draw a parallel $\ell$ to $AB$, and then choose a point $M$ not on $AB$ or on $\ell$. If we put $A'=MB\cap \ell,B'=MA\cap \ell,T=AA'\cap BB'$, then $MT$ cuts the line $AB$ in the midpoint of $AB$.

d). Let $A,B$ be two points on the line we are given and let $K$ be the third point. We can construct the midpoint $P$ of $AB$. Choose a point $M$ and put $A'=AK\cap MB,B'=BK\cap MA$. The line through $K$ parallel to $AB$ is the line passing through $K$ and the harmonic conjugate of $M$ wrt $A',B$. The harmonic conjugate is easily constructed with the ruler alone, by using the harmonic properties of the complete quadrilateral: choose a point $R$ s.t. $RM$ is a diagonal in a complete quadrilateral having $A',B$ as two of its vertices. The line connecting $R$ to the intersection of the other two diagonals cuts $A'B$ in the point we are looking for. If we call this point $X$, we will have $KX\|AB$.

c) How to construct segment bisector. Let $A$ and $B$ to points. We can construct two parallel lines through $A$ and $B$ in two ways. So we obtain rhombus $ACBD$, where $CD$ is a bisector of $AB$. a) and b) are really easy.

Your approach for c) did come to my mind, but it's not that simple because $AB$ might be shorter than the distnce between the two parallel sides of the ruler . I think it can be repaired, but I was tired last night so I just let it go and went to sleep .

Here's a correct version : Put a point $C$ on $(AB$ s.t. $BC>$ than the distance between the sides of the ruler. We can find, with the aid of d), two points $E,F$ not on $AB$ s.t. $EFBC$ is a parallelogram. We can then find a point $D$ on $(BA$ s.t. $EFDA$ is a parallelogram. We now have $AD=BC>$ the distance between the sides of the ruler, and the perpendicular bisector of $DC$ is the same as that of $AB$, so we can perform Myth's construction on the segment $DC$ instead of $AB$, obtaining the perpendicular bisector of $AB$. Perhaps there are simpler ways to do this.

Indeed, there are simple way. Consrtruct parallel line to $AB$. Using rule construct on this line three points $C$, $D$, $E$, $CD=ED$. Let $AC$ and $ED$ intersect in p. $F$, then $FD$ intersects $AB$ in p. $G$ s.t. $AB=BG$! Now I think you are able to complete...

Here's yet another way: construct the perpendicular bisectors of large enough segments $AC,BC$, which must intersect in the circumcenter of $ABC$, $O_C$. The line connecting $O_C$ and the midpoint of $AB$ is the perpendicular bisector of $AB$. I bet we could continue to post proofs all day long, but I see no point in that .