Two point raytracing for reflection off a 3D plane 
186. At Ellipsi in fig. 58 ductis HFN, Hfn, bini FPH, FpH æquales erunt binis fPM, fpm, sive quatuor internis, & oppofitis PfH, PHf, pfH, pHf, nimirum toti PHp, & toti Pfp. Angulus autem PFp æqualis binis PFN, pFN, sive quatuor internis FPH, FHP, FpH, FHp, vel binis illis FPH, FpH cum angulo PHp, adeoque angulo PHp bis, & toti Pfp semel. Quare angulo Pfp dempto a PFp, remanet angulus PHp bis. 

186. In the ellipse in fig. 58, draw HFN and Hfn. Then FPH and FpH are equal to fPM and fpm respectively and so the four internal opposite angles PfH, PHf, pFH, and pfH evidently sum to PHp with Pfp. ^{1}Now angle PFp is the sum of PFN and pFN and so the sum of internal angles FPH, FHP, FpH, and FHp, hence [the sum of] FPH, FpH and angle PHp. [From above,] this is precisely equal to PHp twice combined with Pfp once. Therefore subtracting Pfp from PFp leaves twice the angle PHp. 

Two point raytracing for reflection off a 3D plane 