Exercise:
Let the straight line AB and CD intersect at O.
1, the intersection with p is PF//AB, PE//CD, and straight lines AB and CD intersect at e and f respectively.
Step 2 connect EF
3. Parallel lines intersect P as EF, and intersect AB and CD at G and H..
Then the line GH is the demand.
Prove:
Connect OP and cross EF to m.
Because the quadrilateral PEOF is a parallelogram
So em = FM
Because EF//GH
So em/pg = om/op, FM/ph = om/op.
So em/pg = FM/ph
So pg = ph
The last proof or the second proof is complicated. There is no need for consistency or proportion.
Like this:
PEFH is a parallelogram = >; PH=EF
PFEG is a parallelogram = >; PG=EF
= & gtPG=PF