# Gayle Gilbert

Consider .  Select a point P inside the circle and draw lines AP, BP, and CP extended to their intersections with the opposite sides in points D, E, and F respectively.

We will prove that  for any triangle.

LetŐs construct a line through B parallel to AD and line CF, and letŐs call the intersection point X.  Now, letŐs construct a line through C parallel to AD and line BE, and letŐs call the intersection point Y.

By vertical angles, , and by the Z-principle, .  Therefore, .  Thus, .

Likewise, .  Thus, .

since  (shared in both triangles) and  (corresponding angles).  Thus, .

Likewise, .  Thus, .

Now, letŐs multiply these three equations together:

Conversely, letŐs suppose that D, E, and F are points on BC, AC, and AB respectively and satisfying

Let Q be the intersection point of AD and BE, and let be the intersection point of CQ and AB.  Since AD, BE, and CFŐ are concurrent, then we have

When we combine our equations, we have

Therefore, AD, BE, and CF are concurrent.

QED.

Note: Since we know CevaŐs Theorem must be satisfied for any concurrent point, then we can use this theorem to prove the concurrency of medians (if P is the centroid), the lines of the altitudes (if P is the orthocenter), the bisectors of the angles (if P is the incenter), the perpendicular bisectors of the sides (if P is the circumcenter), or any other special concurrency points.  If we can prove that CevaŐs Theorem is satisfied with any of these special points, then we can prove that the points are indeed concurrent.