+1439 Quadrilateral $PQRS$ is a trapezoid with bases $\overline{PQ}$ and $\overline{RS}$. The median $\overline{MN}$ meets the diagonals $\overline{PR}$ and $\overline{QS}$ at $X$ and $Y$, respectively. If $PS = QR$, then find $RY$.
+129850 Let the intersection of the diagonals = T
The distance from the base to the midline = 5
Triangle PTQ is similar to triangle RTS and both are isosceles with PT =QT and ST = RT
So....the height of each bisect their bases
Since PQ / SR = 30 :20 = 3 : 2.....there are 5 equal parts to the height of the trapezoid and the height of triangle PTQ is 3 of them.....so its height =(3/5)10 = 6 and triangle RTS has a height of 4
And triangle XTY is similar to triangle PTQ
So XTY is also isoceles and its height also bisects its base
The height of triangle XTY = height of triangle PTQ - height of midline = 6 - 5 = 1
So height of triangle XTY to height of triangle PTQ = 1 : 6
So the base of triangle XTY to the base of triangle PTQ = 1 : 6
So the base of triangle XTY = 30 /6 = 5
Let the midpoint of the base of triangle XTY = W
And WY = 1/2 the base of triangle XTY = 5/2 = 2.5
Let W = (0,0)
Let Y =(2.5,0)
Let R = (10,5)
So RY = sqrt [ (10 - 2.5)^2 + 5^2 ] = sqrt[ 7.5^2 + 25] = sqrt [ 56.25 + 25 ] = sqrt [ 81.25] ≈ 9.01
