proof of Van Aubel’s theorem

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

$$\frac{y+z}{x}=\frac{u+v}{w}\mathit{\hspace{1em}\hspace{1em}}\frac{w+x}{v}=\frac{y+z}{u}\mathit{\hspace{1em}\hspace{1em}}\frac{u+v}{z}=\frac{w+x}{y}$$

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

$w(y+z)$ $=$ $x(u+v)$ (1)

$y(u+v)$ $=$ $z(w+x)$ (2)

$u(w+x)$ $=$ $v(y+z)$ (3)

which imply

$$vxz=uwy$$ (4)

and the conclusion says that

$$x(\underset{¯}{wy}+\underset{¯}{wz}+uw+\underset{¯}{xy}+\underset{¯}{xz}+ux+\underset{¯}{y^2}+\underset{¯}{yz}+uy$$

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

equals

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

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), and the proof is complete.

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

Title	proof of Van Aubel’s theorem
Canonical name	ProofOfVanAubelsTheorem
Date of creation	2013-03-22 13:58:00
Last modified on	2013-03-22 13:58:00
Owner	mathcam (2727)
Last modified by	mathcam (2727)
Numerical id	8
Author	mathcam (2727)
Entry type	Proof
Classification	msc 51N20
Related topic	VanAubelTheorem
Related topic	ProofOfVanAubelTheorem