Book II · Proposition 6

II.6

If a straight line be bisected and a straight line be added to it in a straight line, the rectangle contained by the whole with the added straight line and the added straight line together with the square on the half is equal to the square on the straight line made up of the half and the added straight line.Heath, 1908

The companion to II.5, and the other half of Euclid's method for quadratics: ab + ((a-b)/2)^2 = ((a+b)/2)^2 with the sign the other way.

Every step, checked

• B lies between A and Dgiven
• C bisects ABby I.10
• the rectangle contained by AD and DB, with the square on the half, equals the square on CDby II.5, I.46
• and CD is the half together with the added lineby C.N.2

What it needs, and what needs it

Needs: I.10 I.46 II.5

Used by: II.11 III.36 VI.29 XIII.1 XIII.3

Rests on: C.N.1, C.N.2, C.N.3, C.N.4, C.N.5, Def.10, Def.15, Def.22, Def.4, Post.5

Depth: 13 steps of argument above the first principles. Parallel postulate: needed.

What it takes on trust

• a circle-circle intersection is used, though no postulate asserts that two circles meet (×11)continuity
• a line-circle intersection is used, though no postulate asserts that a line meets a circle (×16)continuity
• the argument reads a 'between' fact off the figureorder

The proposition as code

@proposition(
    "II.6",
    THEOREM,
    sample=_bisected_and_produced,
    note="The companion to II.5, and the other half of Euclid's method for "
    "quadratics: ab + ((a-b)/2)^2 = ((a+b)/2)^2 with the sign the other way.",
)
def prop_II_6(a: Point, b: Point, d: Point) -> Out:
    """AB is bisected at C and produced to D."""
    hypothesis("B lies between A and D", between(a, b, d))
    c = posit(prop_I_10(a, b).midpoint, "C")

    lift = _across(b, d)
    whole = _rectangle(a, d, lift, "M", "K")
    outline(a, d, close=False)

    half, added = length(a, c), length(b, d)
    because(prop_II_5, a, b, d)
    because(prop_I_46, a, c)

    claim("C bisects AB", "I.10", eq_len(a, c, c, b))
    claim("the rectangle contained by AD and DB, with the square on the half, "
          "equals the square on CD", ["II.5", "I.46"],
          _area(*whole) + half * half == length(c, d) * length(c, d))
    claim("and CD is the half together with the added line", "C.N.2",
          length(c, d) == half + added)
    return Out(rectangle=whole, midpoint=c)