Question 1062408
Here is a triangle ABC, including points P, Q, R, a few extra midpoints, and a midsegment.
{{{drawing(300,300,-5,5,-1,9,
triangle (-4,0,4,0,2,8),
triangle (-4,0,0,5.333,4,0),
triangle (0,0,4,0,2,8),
green (line (-1,4,3,4)),
locate(2,8.5,A),locate(4,0,B),locate(-4.1,0,C),
locate (-0.1,0,P),locate (1.2,4.5,Q),locate (-0.3,5.8,R),
locate (-1.4,4.25,S),locate (-2.1,-0.1,U),locate (3.2,4.25,T),
red(circle(-2,0,0.05))
)}}} S, T, and U are the midpoints of AC,  AB,  and PC respectively.
Of course, ST is the midsegment of ABC, and SQ is the midsegment of APC.

Since midsegments are half a long as the base, {{{SQ=(1/2)PC=(1/4)BC=PU=UC}}} .
Since SQ and UC are congruent and parallel,
SQUC is a parallelogram,
and QU is parallel to AC .
With their pairs of parallel sides, triangles SQR and UBQ have 3 pairs of congruent angles.
That makes triangles SQR and UBQ similar triangles.
Since {{{UC=(1/4)BC}}} , {{{UB=(3/4)BC}}} .
The corresponding side in SQR is {{{SQ=(1/4)BC}}} ,
So, sides of UBQ are {{{3}}} times longer than corresponding sides of SQR :
UB = 3SQ , UQ = 3SR , and {{{highlight (BQ=3QR)}}} .