Commentary on Proposition XI.15¶
Back to Proposition XI.15.
This result is arrived at in the American text-books already quoted by starting from the relation between a plane and a straight line parallel to it. The series of propositions is worth giving. A straight line and a plane being parallel if they do not meet however far they may be produced, we have the following propositions.
Any plane containing one, and only one, of two parallel straight lines is parallel to the other.
For suppose AB, CD to be parallel and CD to lie in the plane MN. Then AB, CD determine a plane intersecting MN in the straight line CD Thus, if AB meets MN, it must meet it at some point in CD. But this is impossible, since AB is parallel to CD. Therefore AB will not meet the plane MN, and is therefore parallel to it. [This proposition and the proof are in Legendre.]
The following theorems follow as corollaries.
Through a given straight line a plane can be drawn parallel to any other given straight line ; and, if the lines are not parallel, only one such plane can be drawn.
We have simply to draw through any point on the first line a straight line parallel to the second line and then pass a plane through these two intersecting lines. This plane is then, by the above proposition, parallel to the second given straight line.
Through a given point a plane can be drawn parallel to any two straight lines in space ; and, if the latter are not parallel, only one such plane can be drawn.
Here we draw through the point straight lines parallel respectively to the given straight lines and then draw a plane through the lines so drawn.
Next we have the partial converse of the first proposition above.
If a straight line is parallel to a plane, it is also parallel to the intersection of any plane through it with the given plane.
Let AB be parallel to the plane MN, and let any plane through AB intersect MN in CD. Now AB and CD cannot meet, because, if they did, AB would meet the plane MN. And AB, CD are in one plane. Therefore AB, CD are parallel. From this follows as a corollary :
If each of two intersecting straight lines is parallel to a given plane, the plane containing them is parallel to the given plane.
Let AB, AC be parallel to the plane MN. Then, if the plane ABC were to meet the plane MN, the intersection would be parallel both to AB and to AC : which is impossible.
Lastly, we have Euclid’s proposition.
If two straight lines forming an angle are respectively parallel to two other straight lines forming an angle, the plane of the first angle is parallel to the plane of the second.
Let ABC, DEF be the angles formed by straight lines parallel to one another respectively. Then, since AB is parallel to DE, the plane of DEF is parallel to AB [(1) above]. Similarly the plane of DEF is parallel to BC. Hence the plane of DEF is parallel to the plane of ABC [(5)].
Legendre arrives at the result by yet another method. He first proves Eucl. XI. 16 to the effect that, if two parallel planes are cut by a third, the lines of intersection are parallel, and then deduces from this that, if two parallel straight lines are terminated by two parallel planes, the straight lines are equal in length.
(The latter inference is obvious because the plane through the parallels cuts the parallel planes in parallel lines ; which therefore, with the given parallel lines, form a parallelogram.)
Legendre is now in a position to prove Euclid’s proposition XI. 15. If ABC, DEF be the angles, make AB equal to DE, and BC equal to EF, and join CA, FD, BE, CF, AD. Then, as in Eucl. XI. 10, the triangles ABC, DEF are equal in all respects ; and AD, BE, CF are all equal. It is now proved that the planes are parallel by reductio ad absurdum from the last preceding result. For, if the plane ABC is not parallel to the plane DEF, let the plane drawn through B parallel to the plane DEF meet CF, AD in H, G respectively. Then, by the last result BE, HF, GD will all be equal. But BE, CF, AD are all equal : which is impossible. Therefore etc.