Start with a triangle ABC and points A' on BC, B' on CA, C' on AB, so that AA', BB' and CC' are concurrent at D. The barycentric coordinates of D are (x, y, z). Then construct point A'' by reflecting A' in the midpoint of BC (i.e., rotate by 180 degrees). Similarly construct B'' by reflecting B' in the midpoint of CA and C'' by reflecting in the midpoint of AB.
Yes. Since the original barycentric coordinates of D are (x, y, z), then the ratios on the sides are
AC'/C'B = y/x; BA'/A'C = z/y, CB'/B'A = x/z.
But if the points are reflected in the midpoints to get new points, the new ratios are teh reciprocals of the old ones:
AC''/C''B = x/y; BA''/A''C = y/z, CB''/B''A = z/x.
So the product of the 3 ratios is (x/y)(y/z)(z/x) = 1. By Ceva the lines AA'', BB'', CC'' are concurrent.
We can find the new barycentric coordinates (x', y', z') by working on two sides. Assume x' = 1. Then we have
So we can either write [x',y', z'] = (1/((1/x) + (1/y) + (1/z)))[1/x, 1/y, 1/z] or perhaps more simply, [x',y', z'] = (1/(yz+xz+xy))[yz, zx, xy]. In any case, the coordinates are proportional to the reciprocals of (x, y, z).
In the (x, y) plane, A = (0,0), B = (1,0), C= (0,1). Line m is the line with equation (x/a) + (y/b) = 1.
Line AB has equation y = 0, so the intersection with m has (x/a) + 0 = 1, or x = a. The point of intersection C' = (a, 0).
Line AC has equation x = 0, so the intersection with m has 0 + (y/b) = 1, or y = b. The point of intersection B' = (0, b).
Line BC has equation x+y = 1. Since the equation of m is also x + (ay/b) = a, we can subtract to get the y coordinate of the intersection: (ay/b) - y = a - 1, or y(a-b)/b = (a-1), so y = b(a-1)/(a-b). Then x = 1-y = (a - b -ab + b)/(a-b) = a(1-b)/(a-b). The point of intersection A' = (1/(a-b))(a(1-b), b(a-1)).
Note: When a = b, the line m is either equal to or parallel to line BC, so A' is not defined.
The ratios can be computed from either the x or y coordinate (except for the case when the coordinate is constant on all the points of a side, such as the x-coordinate on AC).
AC'/C'B has x-coordinate ratio (a-0)/(1-a) = a/(1-a).
CB'/B'A has y-coordinate ratio (b-1)/(0-b) = (1-b)/b.
BA'/A'C has x-coordinate ratio {(a(1-b)/(a-b)) - 1}/{0 - (a(1-b)/(a-b))}
The numerator is (a-ab-a+b)/(a-b) = (b-ab)/(a-b) = b(1-a)/(a-b) and the denominator is a(b-1)/(a-b)
Thus the ratio BA'/A'C = b(1-a)/a(b-1).
Note: Exceptional Case. In addition to the case a = b above, if a = 1 or b = 1, so that m passes through a vertex of the triangle, then one or more ratio will be infinite and thus not defined.
(AC'/C'B)(CB'/B'A)(BA'/A'C) =(a/(1-a))((1-b)/b)(b(1-a)/a(b-1)).
Collecting numerator and denominator, this is{a(1-b)b(1-a)/{(1-a)ba(b-1) = -1
The ratios in Ceva's theorem that guarantee concurrence of AA', BB', CC' have product +1! In this case the ratio product = -1.
YES. This proves the Theorem of Menelaus. The reason is that for any triangle DEF and line n, the triangle DEF can be mapped to the special triangle ABC by an affine transformation, and the line n will be mapped to some line m. As noted above, there are two exceptional cases: n cannot be parallel to any side of the triangle and also n cannot pass through any of the vertices DEF.
Theorem of Menelaus: Given 3 points A' on line BC, B' on line CA and C' on line AB (each point A', B', C' distinct from a vertex), the points are collinear if and only if the product of the ratios
(AC'/C'B)(CB'/B'A)(BA'/A'C) = -1.
Note: We have clearly proved that if the points are collinear the product = -1. But the converse also follows. Suppose the product of ratios = -1. Then suppose line A'B' intersects AB in point C*. Then the product of ratios with C* instead of C' will also be -1, so equating and canceling, we get AC'/C'B = AC*/C*B. But there is only one point on AB with this ratio, so C' = C* and C' is on line A'B'.