1. extend the extension lines of EM and FN to p and let M(a, b) N(c, d).

MP=c-a MP=b-d

Because both m and n are on the inverse scale image, ab=cd means A/C = D/B..

According to the nature of proportion, (c-a)/c = (b-d)/b.

That is, MP/EP=NP/FP, and the angle P is a common angle, so the triangle PMN is similar to the triangle PEF.

So angel ·PMN = angel ·PEF

MN is parallel to EF.

2. The same truth can be proved.