# 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

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

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), and the proof is complete.

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

Title proof of Van Aubel’s theorem ProofOfVanAubelsTheorem 2013-03-22 13:58:00 2013-03-22 13:58:00 mathcam (2727) mathcam (2727) 8 mathcam (2727) Proof msc 51N20 VanAubelTheorem ProofOfVanAubelTheorem