Commentary on Proposition XI.18¶
Back to Proposition XI.18.
Starting as Euclid does from the definition of perpendicular planes as planes such that all straight lines drawn in one of the planes at right angles to the common section are at right angles to the other plane, it is necessary for him to show that, if F be any point in CE, and FG be drawn in the plane DE at right angles to CE, FG will be perpendicular to the plane to which AB is perpendicular.
It is perhaps more scientific to make the definition, as Legendre makes it, a particular case of the definition of the inclination of planes. Perpendicular planes would thus be planes such that the angle which (when it is acute) Euclid calls the inclination of a plane to a plane is a right angle. When to this is added the fact incidentally proved in XI. 10 that the “inclination of a plane to a plane” is the same at whatever point in their common section it is drawn, it is sufficient to prove the perpendicularity of two planes if one straight line drawn, in one of them, perpendicular to their common section is perpendicular to the other.
If this point of view is taken, Props. 18, 19 are much simplified (cf. Legendre, H. M. Taylor, Smith and Bryant, Rausenberger, Schultze and Sevenoak, Holgate). The alternative proof is as follows.
Let AB be perpendicular to the plane MN, and CE any plane through AB, meeting the plane MN in the straight line CD. In the plane MN draw BF at right angles to CD.
Then ABF is the angle which Euclid calls (in the case where it is acute) the “inclination of the plane to the plane.”
But, since AB is perpendicular to the plane MN, it is perpendicular to BF in it. Therefore the angle ABF is a right angle; whence the plane CE is perpendicular to the plane MN.