Commentary on Proposition XI.4

Back to Proposition XI.4.

The steps to be successively proved in order to establish this proposition by Euclid’s method are

  1. triangles AED, BEC equal in all respects, [by I. 4]

  2. triangles AEG, BEH equal in all respects, [by I. 26] so that AG is equal to BH, and GE to EH,

  3. triangles AEF, BEF equal in all respects, [I. 4] so that AF is equal to BF,

  4. likewise triangles CEF, DEF, so that CF is equal to DF,

  5. triangles FAD, FBC equal in all respects, [I. 8] so that the angles FAG, FBH are equal, (6) triangles FAG, FBH equal in all respects, [by (2), (3), (5) and I. 4] so that FG is equal to FH,

  1. triangles FEG, FEH equal in all respects, [by (2), (6) and I. 8] so that the angles FEG, FEH are equal, and therefore FE is at right angles to GH.

In consequence of the length of the above proof others have been suggested, and the proof which now finds most general acceptance is that of Cauchy, which is as follows.

Let AB be perpendicular to two straight lines BC, BD in the plane MN at their point of intersection B. In the plane MN draw BE, any straight line through B. Join CD, and let CD meet BE in E. Produce AB to F so that BF is equal to AB. Join AC, AE, AD, CF, EF, DF.

A M B D E C N F

Since BC is perpendicular to AF at its middle point B,

AC is equal to CF.
Similarly AD is equal to DF.
Since in the triangles ACD, FCD the two sides AC, CD are respectively equal to the two sides FC, CD, and the third sides AD, FD are also equal,
[I. 8]the angles ACD, FCD are equal.

The triangles ACE, FCE thus have two sides and the included angle equal, whence

[I. 4]EA is equal to EF.

The triangles ABE, FBE have now all their sides equal respectively;

therefore the angles ABE, FBE are equal, [I. 8] and AB is perpendicular to BE.

And BE is in any straight line through B in the plane MN.

Legendre’s proof is not so easy, but it is interesting. We are first required to draw through any point E within the angle CBD a straight line CD bisected at E. To do this we draw EK parallel to DB meeting BC in K, and then mark off KC equal to BK. CE is then joined and produced to D; and CD is the straight line required.

B D K E C

Now, joining AC, AE, AD in the figure above, we have, since CD is bisected at E,

  1. in the triangle ACD,

AC2 $+$ AD2 $= 2$ AE2 $+ 2$ ED2,

and also (2) in the triangle BCD,

BC2 $+$ BD2 $= 2$ BE2 $+ 2$ ED2.

Subtracting, and remembering that the triangles ABC, ABD are right-angled, so that

AC2 $-$ BC2 $=$ AB2,

and

AD2 $-$ BD2 $=$ AB2,

we have

$2$ AB2 $= 2$ AE2 $- 2$ BE2,

or

AE2 $=$ AB2 $+$ BE2,
whence [I. 48] the angle ABE is a right angle, and AB is perpendicular to BE.

It follows of course from this proposition that the perpendicular AB is the shortest distance from A to the plane MN.

And it can readily be proved that, If from a point without a plane oblique straight lines be drawn to the plane, (1) those meeting the plane at equal distances from the foot of the perpendicular are equal, and (2) of two straight lines meeting the plane at unequal distances from the foot of the perpendicular, the more remote is the greater.

Lastly, it is easily seen that From a point outside a plane only one perpendicular can be drawn to that plane.

For, if possible, let there be two perpendiculars. Then a plane can be drawn through them, and this will cut the original plane in a straight line.

This straight line and the two perpendiculars will form a plane triangle which has two right angles: which is impossible.