Commentary on Proposition XI.8

Back to Proposition XI.8.

Simson objects to the words which explain why DC is in the plane through BD, DA, viz. “inasmuch as AB, BD are in the plane through BD, DA, and DC is also in the plane in which AB, BD are,” as being too roundabout. He concludes that they are corrupt or interpolated, and that we ought only to have the words “because all three are in the plane in which are the parallels AB, CD” (by Prop. 7 preceding). But I think Euclid’s words can be defended. Prop. 7 says nothing of a plane determined by two transversals as BD, DA are. Hence it is natural to say that DC is in the same plane in which AB, BD are [Prop. 7], and AB, BD are in the same plane as BD, DA [Prop. 2], so that DC is in the plane through BD, DA.

Legendre’s alternative proof is split by him into two propositions.

  1. Let AB be a perpendicular to the plane MN and EF a line situated in that plane; if from B, the foot of the perpendicular, BD be drawn perpendicular to EF, and AD be joined, I say that AD will be perpendicular to EF.

  2. If AB is perpendicular to the plane MN, every straight line CD parallel to AB will be perpendicular to the same plane.

To prove both propositions together we suppose CD given, join BD, and draw EF perpendicular to BD in the plane MN.

A C M B F D E N

  1. As before, we make DE equal to DF and join BE, BF, AE, AF.

Then, since the angles BDE, BDF are right, and DE, DF equal,
[I. 4]BE is equal to BF.
And, since AB is perpendicular to the plane,
the angles ABE, ABF are both right.
Therefore, in the triangles ABE, ABF, AE is equal to AF. [I. 4]
Lastly, in the triangles ADE, ADF, since AE is equal to AF, and DE to DF, while AD is common,
the angle ADE is equal to the angle ADF, [I. 8] so that AD is perpendicular to EF.

  1. ED being thus perpendicular to DA, and also (by construction) perpendicular to DB,

[XI. 4]ED is perpendicular to the plane ADB.
But CD, being parallel to AB, is in the plane ADB;
[XI. Def. 3]therefore ED is perpendicular to CD.
Also, since AB, CD are parallel, and ABD is a right angle,
CDB is also a right angle. Thus CD is perpendicular to both DE and DB, and therefore to the plane MN through DE, DB.