+128578 3. By a property of medians ON = (1/3)QN = (1/3)(12) = 4
NR = PR/2 = 14 / 2 = 7
And triangle ONR is right such that
sqrt [ NR^2 + ON^2 ] = sqrt [ 7^2 + 4^2 ] = sqrt [ 65] = OR
+128578 4. 
Semi-perimeter of PQR = [ 17 + 10 + 9 ] / 2 = 18
Area of PQR = sqrt [ 18 * 1 * 8 * 9 ] = sqrt [ 1296 ] = 36
Find altitude SQ
36 = (1/2)PR * SQ
72 = 17 * SQ
SQ = 72 / 17
Let XR = x Let QX = (10 - x)
By an angle bisector property
XR / PR = QX / PQ
x / 17 = (10 - x) / 9
9x = 17 (10 - x)
9x = 170 - 17x
26x = 170
x = 170 / 26 = 85 / 13 = XR
Triangles XYR and QSR are similar
XY /XR = QS / QR
XY / (85/13) = (72/17) /10
XY = (85/13)(72/17) / 10 = 36 / 13
+128578 2. H is the orthocenter of the triangle Draw CH and extend it to F where CF is the third altitude of the triangle
Angle AHE = 180 - 128 = 52
Angle AEH = 90
Angle HAE = 90 - 52 = 38
Angle BAH = 28
Angle BAC = Angle BAH + Angle HAE = 28 + 38 = 66
Angle CFA = 90
Angle BAC = 66
Angle FCA = 180 - 90 - 66 = 24
But Angle FCA = Angle HCA = 24
