Plane Affine Algebra (III)

See also Part I and Part II.

Finally, a theorem, in the style of the Elements:

To trisect a segment. Let the endpoints of the segment be P, Q. Let R be any point not on PQ. Let M be the midpoint of PR, N be the midpoint of QR, O the midpoint of PQ. Then let S be the midpoint of MO, and T be the midpoint of NO. Produce the line RS to PQ in X; similarly, produce the line RT to PQ in Y. Then PX, XY, YZ are equal to one-third of PQ.

Proof: We begin by calculating S and T:

   S
= { definition with M, O := ray.P.R.½, ray.P.Q.½ }
   ray.(ray.P.R.½).(ray.P.Q.½).½
= { (18), twice }
   ray.(P → ½·‹R−P›).(P → ½·‹Q−P›).½
= { (18) }
   P → ½·‹R−P› → ½·‹(P → ½·‹Q−P›)−(P → ½·‹R−P›)›
= { (3); linearity }
   P → ½·(‹R−P› + ‹(P → ½·‹Q−P›)−(P → ½·‹R−P›)›)
= { (9) }
   P → ½·(‹R−P› + ‹P−P› + ½·‹Q−P› − ½·‹R−P›)
= { (6); linearity }
   P → ¼·(‹R−P› + ‹Q−P›)

From considerations of symmetry, T = S[P, Q := Q, P]. We have:

(21)  S = P → ¼·(‹R−P› + ‹Q−P›)
(22)  T = Q → ¼·(‹R−Q› + ‹P−Q›)

Finally, we calculate X and Y:

 X
= { definition }
 meet.P.Q.R.S
= { (20); (18) }
 P → (‹S−R›⊥⋅‹R−P›)/(‹S−R›⊥⋅‹Q−P›)·‹Q−P›
= { (21) }
 P → (‹(P → ¼·(‹R−P› + ‹Q−P›))−R›⊥⋅‹R−P›)/(‹(P → ¼·(‹R−P› + ‹Q−P›))−R›⊥⋅‹Q−P›)·‹Q−P›
= { (2), twice }
 P → ((‹P−R› + ¼·(‹R−P› + ‹Q−P›))⊥⋅‹R−P›)/((‹P−R› + ¼·(‹R−P› + ‹Q−P›))⊥⋅‹Q−P›)·‹Q−P›
= { (5), twice; linearity throughout }
 P → ((¼·‹Q−P› − ¾·‹R−P›)⊥⋅‹R−P›)/((¼·‹Q−P› − ¾·‹R−P›)⊥⋅‹Q−P›)·‹Q−P›
= { (10) and (11), twice }
 P → ((¼·‹Q−P›⊥ − ¾·‹R−P›⊥)⋅‹R−P›)/((¼·‹Q−P›⊥ − ¾·‹R−P›⊥)⋅‹Q−P›)·‹Q−P›
= { linearity of inner product, twice }
 P → (¼·‹Q−P›⊥⋅‹R−P› − ¾·‹R−P›⊥⋅‹R−P›)/(¼·‹Q−P›⊥·‹Q−P› − ¾·‹R−P›⊥⋅‹Q−P›)·‹Q−P›
= { (14), twice }
 P → (¼·‹Q−P›⊥⋅‹R−P›)/(−¾·‹R−P›⊥⋅‹Q−P›)·‹Q−P›
= { (12) }
 P → (¼·‹Q−P›⊥⋅‹R−P›)/(¾·‹Q−P›⊥⋅‹R−P›)·‹Q−P›
= { algebra }
 P → ⅓·‹Q−P›

To finalize, from considerations of symmetry, Y = meet.Q.P.R.T = (meet.P.Q.R.S)[P, Q := Q, P]. Hence:

(23)  X = P → ⅓·‹Q−P›
(24)  Y = Q → ⅓·‹P−Q›