Showing existence of line which cuts portions equally and point of intersections being collinear from $ZP,ZQ,ZR,ZS$ lines to line A,B,C,D respectively

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Consider four points $P,Q,R,S$ on a line such that $PQ = m, QR= n , RS= o$ , $m \neq n \neq o $. Let $X$ be a point not on the line such that lines $L_1,L_2,L_3,L_4$ are constructed as shown in figure : enter image description here

  • $1)$ Show that there exist many lines $l$ which cuts the other branch of pencil of lines from X in equal portion (i.e. $P'Q': Q'R' :R'S'= 1:1:1$) while varying X to any position . If given that there exists a line $l_°$ in which there exists $P'_°Q'_° :Q'_°R'_°:R'_°S'_°= 1:1:1$ for some specific $X$ position ( that is just one existence shows thats its always possible for other $X$ positions )

    Now consider making the lines $A,B,C,D$ pependicular to one of possible line $l$ and passing through each point of intersection : $P',Q'R',S'$ . Let $Z$ be point of feet of perpendicular from $X$ to line $l$ , let $ZP$ line intersects line $A$ at $a$, simiarily others like $ZQ$ intersecting line $B$ at $b$ ,

  • $2)$ Show that the points $a,b,c,d$ are collinear .

Note: i was able to realize this from a physics problem, this all came into use there, but was not able to prove it mathematically so asked this. My progress: i tried using some similar triangles to make a possible line but fails :(