Book VI · Proposition 29

VI.29

To a given straight line to apply a parallelogram equal to a given rectilineal figure and exceeding by a parallelogrammic figure similar to a given one.Heath, 1908

The other root, and the other sign of the quadratic: here the figure runs past the end of the line instead of falling short of it.

Every step, checked

• A and B are distinctgiven
• the application is a proper onegiven
• the point falls beyond B on AB producedby II.6
• the rectangle applied equals the given areaby II.6
• and it exceeds AB by the figure on BSby II.6

What it needs, and what needs it

Needs: I.10 II.6

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: 14 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 (×8)continuity
• a line-circle intersection is used, though no postulate asserts that a line meets a circle (×8)continuity

The proposition as code

@proposition(
    "VI.29",
    CONSTRUCTION,
    sample=_deficient_application,
    note="The other root, and the other sign of the quadratic: here the figure "
    "runs past the end of the line instead of falling short of it.",
)
def prop_VI_29(a: Point, b: Point, part) -> Out:
    """Apply to AB a parallelogram equal to a given area, exceeding by a square."""
    hypothesis("A and B are distinct", a != b)
    hypothesis("the application is a proper one", sign(part) > 0 and sign(1 - part) > 0, guard=True)
    line(a, b, "the given line AB")
    middle = posit(prop_I_10(a, b).midpoint, "C")

    whole = length(a, b)
    wanted = whole * whole * part * (1 + part)
    # x(x - whole) = wanted always has a root, however large the area: this is
    # the case II.6 covers, and it needs no proviso at all.
    excess = sqrt(whole * whole / 4 + wanted)
    beyond = posit(_along(a, b, (whole / 2 + excess) / whole), "S")
    across = (-(b.y - a.y), b.x - a.x)
    applied = _parallelogram_on(a, beyond, Point(a.x + across[0], a.y + across[1]))
    outline(*applied)

    because(prop_II_6, a, b, beyond)

    claim("the point falls beyond B on AB produced", "II.6",
          on_line(beyond, Line.through(a, b))
          and sign(length(a, beyond) - length(a, b)) > 0)
    claim("the rectangle applied equals the given area", "II.6",
          length(a, beyond) * length(b, beyond) == wanted)
    claim("and it exceeds AB by the figure on BS", "II.6",
          length(a, beyond) == whole + length(b, beyond))
    return Out(section=beyond, midpoint=middle)