Hello ordinalhigh

I have a proof that does indeed use Ceva's Theorem. Here's the outline.

Look at the attached diagram. We have to prove that:

$\displaystyle \frac{AP}{AL}+\frac{BP}{BM}+\frac{CP}{CN}= 2$

Let $\displaystyle L,\; M,\; N$ divide $\displaystyle BC,\; CA,\; AB$ in the ratios $\displaystyle p:1,\; q:1,\; r:1$. Then, as shown in the diagram:

$\displaystyle AN = \frac{rc}{1+r}$

$\displaystyle NB = \frac{c}{1+r}$

... etc

Ceva's Theorem states that:$\displaystyle \frac{AN}{NB}\cdot\frac{BL}{LC}\cdot\frac{CM}{MA}= 1$

$\displaystyle \Rightarrow pqr=1$ ...(1)

Next, we construct the line $\displaystyle LQ \parallel BA$, meeting $\displaystyle CN$ at $\displaystyle Q$ and, using similar triangles, we prove that:$\displaystyle LQ = \frac{BN.LC}{BC}=\frac{AN.PL}{AP}$

and hence that$\displaystyle \frac{PL}{AP}= \frac{BN.LC}{AN.BC}$

$\displaystyle \Rightarrow 1+\frac{PL}{AP}= 1+\frac{BN.LC}{AN.BC}$

$\displaystyle \Rightarrow \frac{AL}{AP}=\frac{AN.BC+BN.LC}{AN.BC}$$\displaystyle = ...$ etc

$\displaystyle = \frac{1+r+rp}{r+rp}$

$\displaystyle \Rightarrow \frac{AP}{AL}=\frac{r+rp}{1+r+rp}$

Similarly:$\displaystyle \frac{BP}{BM}=\frac{p+pq}{1+p+q}$

and:$\displaystyle \frac{CP}{CN}=\frac{q+qr}{1+q+qr}$

Adding these results together:$\displaystyle \frac{AP}{AL}+\frac{BP}{BM}+\frac{CP}{CN}=\frac{r+ rp}{1+r+rp}+\frac{p+pq}{1+p+pq}+\frac{q+qr}{1+q+qr }$

From (1):$\displaystyle r = \frac{1}{pq}$

$\displaystyle \Rightarrow \frac{AP}{AL}+\frac{BP}{BM}+\frac{CP}{CN}=\frac{\d frac{1}{pq}+\dfrac{1}{q}}{1+\frac{1}{pq}+\frac{1}{ p}}+\frac{p+pq}{1+p+pq}+\frac{q+\dfrac{1}{p}}{1+q+ \dfrac{1}{p}}$

$\displaystyle = ...$ etc

$\displaystyle =2$

I'm sure you can fill in the gaps.

Grandad