PLEASE I URGENTLY NEED YOUR HELP These problems are from my

PLEASE! I URGENTLY NEED YOUR HELP! These problems are from my Advanced Geometry class \"Principles of Geometry\":

In the figure below. Suppose that XY||AB and YZ||BC. Prove that XZ||AC. In the figure below. Suppose that PY||AQ and QX||BP. Prove that XY||AB.

Solution

Q.7
We know that Dist(A,X) = Dist(B,Y) (Given that XY is parallel to AB)
   We also know that Dist(B,Y) = Dist(C,Z) (Given that BC is parallel to YZ)

Thus, from the two statements, Dist(A,X) = Dist(C,Z) => AC is always at same distance from XZ
=> So by definition, AC and XZ are parallel.

Q.8
   First of all, I think there is a slight printing error in the question, Point Q is incorrectly labeled as Point Q, that might be the reason that you had this doubt in the first place.So i think you can solve it now (i\'m still giving you the solution just in case you can\'t solve it. But please try it yourself once before looking at the solution, best of luck.)

In CXZ and CPB, CXY = CPB (Correspoonding Angles)
  CZX = CBP (Correspoonding Angles)
=>CXY ~ CPB (By AA rule)
=> CX / CP = CZ / CB (By CPST) -------------------- 1.

Similarly, CPY and CAZ, CP / CA = CY / CZ -------------------- 2.

Multiplying 1. and 2., we get
  CX / CA = CY / CB
   Also, C = C (As it is common to both CXY and CAB)
   =>CXY ~ CAB (By SAS rule)
=>CXY = CAB and CYX = CBA (By CPST)
Therefore, taking corresponding angles to be equal and AX and BY as transversals respectively, l(AB) is parallel to l(XY).