A corollary of Ceva's theorem? Say we have a triangle $ABC$ with three cevians $AM$, $BN$ and $CK$ which intersect at point $O$ inside a triangle. By Ceva's theorem, we know $\frac{ |AK |}{|KB|} \frac{ |BM|}{|MC|} \frac{ |CN|}{|NA|}  = 1$. How does this imply that
$$ \frac{ |OM| }{|AM | } + \frac{ |ON|}{|BN| } + \frac{ |OK|}{|CK|} = 1 $$and
$$ \frac{|OA|}{|AM|} + \frac{ |OB|}{|BN|} + \frac{ |OC|}{|CK|} = 2\qquad ??$$
I mean, in my books it says it follows easily from Ceva's theorem, but I am unable to see why. Do I need to use a barycentric argument?
 A: There is a very simple explanation for the first equality.
Let $(a,b,c)$ be the barycentric coordinates of $O$, meaning that $O=aA+bC+cC$ with
$$a+b+c=1 \ \ \ (*)$$
The first equation $\frac{ |OM| }{|AM | } + \frac{ |ON|}{|BN| } + \frac{ |OK|}{|CK|} = 1$ simply expresses relationship (*) !
Let us understand it on the first barycentric coordinate $a$:
$$\frac{OM}{AM}=a=\frac{\text{area}(OBC)}{\text{area}(ABC)}$$
because the ratio of areas of 2 triangles having a common base is equal to the ratio of their altitudes. It is sometimes called the areal interpretation of barycentric coordinates. See
https://en.wikipedia.org/wiki/Barycentric_coordinate_system
The second equality is easily obtained from the first one by writing its left hand side so:
$$\frac{ |OM|+|MA| }{|MA | } + \frac{  |ON|+|NB|}{|NB| } + \frac{ |OK|+|KC|}{|KC|}.$$
A: How straightforward an answer is depends largely on how you approached Ceva's theorem.
In a Cartesian approach, noting that Affine maps preserve ratios, we  send $A, B, C$ to $A' = (0,1), B' = (0,0), C' = (1,0)$ and prove the relation in the image.
Setting $O' = (u,v)$, we deduce the coordinates of $M', K', N'$. For instance, the line $A'O'$ has slope $\frac{y-y_{A'}}{x-x_{A'}} = \frac{1-v}{0-u}$, and intersects $B'C'$ which has $y =  0$, to give the point $$M' = (\frac{u}{1-v}, 0)$$
Similarly, $$N' = (\frac{u}{u+v}, \frac{v}{u+v})\\ K' = (0, \frac{u}{1-v})$$
So, $$\frac{|O'M'|}{|A'M'|} = \frac{y_{M'}-y_{O'}}{y_{A'}-y_{'M'}} = \frac{v-0}{1-0} = v$$
$$\frac{|O'N'|}{|B'N'|} = \frac{\frac{u}{u+v}-u}{\frac{u}{u+v}}= 1-u-v$$
$$\frac{|O'K'|}{|C'K'|} = \frac{u-0}{1-0} = u$$ and their sum is clearly $1$. 
