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 `P``Q`. Let `M` be the midpoint of `P``R`, `N` be the midpoint of `Q``R`, `O` the midpoint of `P``Q`. Then let `S` be the midpoint of `M``O`, and `T` be the midpoint of `N``O`. Produce the line `R``S` to `P``Q` in `X`; similarly, produce the line `R``T` to `P``Q` in `Y`. Then `P``X`, `X``Y`, `Y``Z` are equal to one-third of `P``Q`.

Proof:We begin by calculatingSandT:S= { definition withM,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

XandY: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›

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