Question 1074278
Maybe by now you have been able to think of a way.
I could not think of a way late last night,
but I woke up with a clearer mind this morning,
so I added line MW, parallel to ST:
{{{drawing(300,300,-1,11,-1,11,
green(triangle(1,5,3,5,1.25,6.25)),
triangle(1.25,6.25,10,0,2,10),
triangle(0,0,4,0,2,10),
line(0,0,10,0),line(0,0,2,10),
line(2,10,10,0),locate(-0.2,0,S),
locate(3.9,0,U),locate(9.9,0,T),
locate(1.9,10.6,R),locate(3.1,5.5,M),
locate(0.9,6.8,V),locate(0.5,5.3,W)
)}}} MW is part of two new triangles similar to two triangles that were there before.
 
Now that I look at the drawing,
I realize it has TU:RS = 3:2 .
A drawing is not necessary to support the reasoning,
but even an inaccurate drawing helps
as a reminder of the meaning of each point.
 
RWM is similar to RSU.
MVW is similar to TVS.
The "scale down factors" between the similar triangles
lead to a way to calculate the lengths of RW and VW
as fractions of the length pf RS,
and that allows to calculate, as a fraction of the length of RS,
the lengths of {{{RV=RW-VW}}} .
 
Since {{{TU+US=TS}}} and {{{TU:US = 2:3}}} ,
{{{TU:US:TS = 2:3:3+2=5}}}
So,
{{{TU=(2/5)TS}}} and {{{US=(3/5)TS}}} .
Since M is the midpoint of RU,
and WM is parallel to ST,
RWM is not only similar to RSU;
it is a {{{1/2}}} scale version of RSU.
So, not only is MW the midsegment of RSU,
with {{{MW=(1/2)US}}} ,
but also {{{RW=(1/2)RS}}} ,
and of course, {{{WS=(1/2)RS}}} too.
  
{{{MW=(1/2)US=(1/2)(3/5)TS=(3/10)TS}}} .
Since MVW is similar to TVS,
and {{{MW=(3/10)TS}}} , {{{VW=(3/10)VS}}} .
 
Finally,
{{{system(VS=VW+WS=VW+(1/2)RS,VW=(3/10)VS)}}} --> {{{VS=(3/10)VS+(1/2)RS}}} .
Form there,
{{{VS-(3/10)VS=(1/2)RS}}} --> {{{(7/10)VS=(1/2)RS}}} --> {{{VS=(10/7)(1/2)RS}}} --> {{{VS=(5/7)RS}}} ,
and {{{RV=RS-VS}}} --> {{{RV=RS-(5/7)RS}}} --> {{{highlight(RV=(2/7)RS)}}}
 
NOTE:
I am not sure if the reasoning above is the shortest way to the answer.
Maybe I should have drawn a line through M parallel to RS instead.
With nothing else to do, I would explore other angles to this problem
(and draw a better drawing).
However, if you are still looking for a geometry-based answer,
a less than perfect answer is better than no answer.