I can do this; but I use coordinate geometry.
Place this diagram on a coordinate axis: Place Q at the origin, QR on the positive x-axis and QP on the
Q = (0,0) P = (0,20) R = (a,0) [use a variable like 'a' because we don't know exactly where it is]
Since RT : TS = 1 : 2, RT will be one-third of SR.
Since SR = 20, RT = 6 2/3.
Coordinates of T = (a, 6 2/3) S = (a,20)
Equation of QS: slope = (20 - 0) / (a - 0) = 20/a y-intercept = 0 ---> y = (20/a)·x
Equation of PT: slope = ( 6 2/3 - 20) / (a - 0)
[multiplying the numerator and the denominator of the slope by 3]
slope = (20 - 60) / (3a) = -40/(3a) y-intercept = 20
---> y = [ -40/(3a) ]·x + 20
Finding where these two lines intersect: (20/a)·x = [ -40/(3a) ]·x + 20
[multiply each term by 3a] 60x = -40x + 60a
100x = 60a
x = (3/5)a
Substituting this value of x into the equation: y = (20/a)·x ---> y = (20/a)·(3/5)a = 12