Cevas theorem without proportions
- Ceva's theorem states that is A', B', C' are points on sidelines BC, CA, AB of triangle ABC then AA', BB', CC' concur if and only if
AC'/ C'B * BA'/A'C *AB'/B'C = 1
One of the possible ways of avoiding segments ratios is to rephrase the theorem in this equivalent form:
If from a triangle vertex, say A, you draw parallel lines to A'C', A'B' and call B'', C'' their intersections with BC; then AA', BB', CC' concur if and only if A' is the middle point of B''C''.
The two statements are equivalent since Ceva's condition can be rewritten this way: A'B*AC'/C'B = CA'* B'A/CB'
but from triangle ABC'' with A'C'//C''A you get C''A'= A'B*AC'/C'B
and from triangle ACB'' with A'B'//B''A you get A'B''= CA'*B'A/CB'
so Ceva condition is equivalent to C''A'=A'B'', that is A' is the midpoint of B''C''. (Sorry if this was already known)
An elementary way (purely affine) to prove this different statement of Ceva's theorem is the following:
Let's suppose we know that A'B''=C''A' and we have to prove that AA', BB', CC' concur.
Call X, Y the points where AB'', AC'' meet BB', CC' and U, V the points where parallels from C'', B'' to AB'', AC'' meet AC and AB and Z their intersection point.
Now consider parallelogram AB''ZC'': since A' is the midpoint of diagonal B''C'' then line AA' passes through Z.
Triangles ZUV and A'B'C' are perspective from A so, by Desargues' affine theorem, UV//B'C'.
Triangles B''XV and A'B'C' are perspective from B so, by Desargues' affine theorem, XV//B'C'.
Triangles C''UY and A'B'C' are perspective from C so, by Desargues' affine theorem, UY//B'C'.
So X, U, V, Y are all on a same line, parallel to B'C', so XY//B'C'.
Triangles AXY and A'B'C' have parallel sides so, by Desargues' affine theorem, they are perspective, that is AA', XB' (BB'), YC' (CC') concur (or are parallel).
The proof of the converse proposition is similar.