Commentary on Proposition XI.11¶
Back to Proposition XI.11.
The text-books differ in the form which they give to this proposition rather than in substance. They commonly assume the construction of a plane through the point A at right angles to any straight line BC in the given plane (the construction being effected in the manner shown at the end of the note on XI. 5 above). The advantage of this method is that it enables a perpendicular to be drawn from a point in the plane also, by the same construction. (Where the letters for the two figures differ, those referring to the second figure are put in brackets.)
We can include the construction of the plane through A perpendicular to BC, and make the whole into one proposition, thus.
BC being any straight line in the given plane MN, draw AD perpendicular to BC. In any plane passing through BC but not through A draw DE at right angles to BC. Through DA, DE draw a plane; this will intersect the given plane MN in a straight line, as FD (AD). In the plane AG draw AH perpendicular to FG (AD). Then AH is the perpendicular required.
In the plane MN, through H in the first figure and A in the second, draw KL parallel to BC. Now, since BC is perpendicular to both DA and DE, BC is perpendicular to the plane AG. [XI. 4] Therefore KL, being parallel to BC, is also perpendicular to the plane AG [XI. 8], and therefore to AH which meets it and is in that plane. Therefore AH is perpendicular to both FD (AD) and KL at their point of intersection. Therefore AH is perpendicular to the plane MN.
Thus we have solved the problem in XI. 12 as well as that in XI. 11; and this direct method of drawing a perpendicular to a plane from a point in it is obviously preferable to Euclid’s method by which the construction of a perpendicular to a plane from a point without it is assumed, and a line is merely drawn from a point in the plane parallel to the perpendicular obtained in XI. 11.