proof of Van Aubel’s theorem

As in the figure, let us denote by $u, v, w, x, y, z$ the areas of thesix component triangles.Given any two triangles of the same height, their areas are in thesame proportion as their bases (Euclid VI.1). Therefore

\\frac{y+z}{x}=\\frac{u+v}{w}\\qquad\\frac{w+x}{v}=\\frac{y+z}{u}\\qquad\\frac{u+v}{z%}=\\frac{w+x}{y}

and the conclusion we want is

\\frac{y+z+u}{v+w+x}+\\frac{z+u+v}{w+x+y}=\\frac{y+z}{x}\\;.

Clearing the denominators, the hypotheses are

$\\displaystyle w(y+z)$	$\\displaystyle=$	$\\displaystyle x(u+v)$	(1)
$\\displaystyle y(u+v)$	$\\displaystyle=$	$\\displaystyle z(w+x)$	(2)
$\\displaystyle u(w+x)$	$\\displaystyle=$	$\\displaystyle v(y+z)$	(3)

which imply

vxz=uwy

(4)

and the conclusion says that

x(\\underline{wy}+\\underline{wz}+uw+\\underline{xy}+\\underline{xz}+ux+\\underline%{y^{2}}+\\underline{yz}+uy

+vz+uv+v^{2}+wz+uw+vw+xz+ux+vx)

equals

(y+z)(vw+vx+vy+w^{2}+wx+wy+\\underline{wx}+\\underline{x^{2}}+\\underline{xy})

or equivalently (after cancelling the underlined terms)

x(uw+xz+ux+uy+vz+uv+v^{2}+wz+uw+vw+ux+vx)

equals

(y+z)(vw+vx+vy+w^{2}+wx+wy)=(y+z)(v+w)(w+x+y)\\;.

i.e.

x(u+v)(v+w+x)+x(xz+ux+uy+vz+wz+uw)=

(y+z)w(v+w+x)+(y+z)(vx+vy+wy)

i.e. by (1)

x(xz+ux+uy+vz+wz+uw)=(y+z)(vx+vy+wy)

i.e. by (3)

x(xz+uy+vz+wz)=(y+z)(vy+wy)\\;.

Using (4), we are down to

x^{2}z+xuy+uwy+xwz=(y+z)y(v+w)

i.e. by (3)

x^{2}z+vy(y+z)+xwz=(y+z)y(v+w)

i.e.

xz(x+w)=(y+z)yw\\;.

But in view of (2), this is the same as (4), andthe proof is complete.

Remarks: Ceva’s theorem is an easy consequence of (4).