- #1

Hunus

- 15

- 0

The proposition, "To a given straight line to apply, in a given rectilineal angle, a parallelogram equal to a given triangle" seemed to me to be proven with, "Let the parallelogram BEFG be constructed equal to the triangle C, in the angle EBG which is equal to the rectilineal angle D [I.42]", but he goes on to employ I.43 (complements of a parallelogram about its diameter are equal to one another) to be able to change the dimensions of the area, but to hold the area constant.

Now I understand that that is important in and of itself, but I don't see why using [I.42] isn't 'applying' the area to the straight line.