I sat my HSC last year but looking at this question would this be valid:
let angle QBA = x
therfore angle BAP = x (alternate angles in parallel lines...)
Similarly, let angle BQP = y
therfore angle APQ = y (alternate angles in parallel lines...)
Now, angle QTA = x + y ( the exterior triangle BQT is equal to the sum of the opposite interior angles)
Also, angle ATP = 180 - (x + y) ( angle sum of triangle PAT 180)
Therefore angle QTP = x + y + 180 - (x +y)
= 180
Thus Q,T,P are collinear
let angle QBA = x
therfore angle BAP = x (alternate angles in parallel lines...)
Similarly, let angle BQP = y
therfore angle APQ = y (alternate angles in parallel lines...)
Now, angle QTA = x + y ( the exterior triangle BQT is equal to the sum of the opposite interior angles)
Also, angle ATP = 180 - (x + y) ( angle sum of triangle PAT 180)
Therefore angle QTP = x + y + 180 - (x +y)
= 180
Thus Q,T,P are collinear