I am trying to work through Book 1 of Apollonius' Conics and got stuck at proposition 1.46 (20 in Heath) where it begins with the assertion that the area of the parallelogram along the tangent of a parabola is equal to the triangle along the ordinate that supersedes it in height. Apparently this follows from proposition 1.42 that demonstrates that a triangle along the tangent is equal to a parallelogram of the same height with half the base. Yet I find this triangle in Proposition 1.46 altogether differently situated then in 1.42--for it extends exactly twice the length of the distance between itself and the parabola, whereas the triangle in 1.46 extends well beyond. Moreover, I don't see how the assertion follows at all. Has Apollonius made an error or I am overlooking something?
I think I understood (from your comment) a possible source of confusion.
In the proof of I.46, $△FRM = △F'R'M$ only means that those triangles have the same area. Apollonius (and Euclid before him) used "equal" with that meaning, when dealing with polygons.
To prove they have the same area, we prove first that $△=⏥()$ and $△R'U'W'=⏥(EW')$ (from I.42) and by subtraction we then get: $$ △RUW- △R'U'W'= ⏥(EW) - ⏥(EW'), $$ that is: $$ quadrilateral(RWW'R')= ⏥(F'W). $$ The pentagon $R'W'WFM$ is common to both quadrilaterals in the above equality. Subtracting it from both sides we finally get: $$ △FRM = △F'R'M. $$
Now one has to prove that $△FRM$ and $△F'R'M$ are congruent. This is true because those triangles are also similar: from $R'F'\parallel RF$ it follwows they have the same angles.
Hence we may conclude that $RM=MR'$.