Carisa Lindsay

Final Project

Ceva’s Theorem

For any triangle ABC,
and any point P within the boundaries of ABC, we can see rather informally that
the ratio of the products of segments created by extending the lines is
one.

Regardless of where
point P is within ABC, the ratio of the products of segment lengths remains a
value of one.

Suppose we didn’t
have access to numeric values and need to see that Ceva’s
Theorem is true. We can prove Ceva’s Theorem through the use of similar triangles.

First, we can see
that AD, BE, and CF are concurrent at point P.

To construct similar
triangles, we first need to construct a transversal (a parallel line) through
point A.

Then we can construct
a similar triangle by extending BE until it intersects with the parallel
line. From this construction, we can build
similar triangles.

Triangle AEG is a
similar triangle to triangle BCE because BG acts as a transversal and we can
use alternate and corresponding angles to show that they are similar.

For similar reasons,
triangle AFH is similar to ABE.

Since these triangles
are similar, we know the following ratios are true:

_{} and _{}

Because these
triangles are similar, we can see from these ratios that their products are
equal.

If we consider the
products of these ratios, we can see that

_{}

Furthermore, we can
also see from the other pair of similar triangles that:

_{} and _{}

Therefore,

_{}

So, if we multiply
the ratios, we will get a result of 1.

_{}

What happens when
point P is outside of ABC?

If point P moves
beyond ABC, we no longer have smaller triangles formed within ABC. Rather, we have only two triangles based on
two of the remaining line segments. For
instance:

Furthermore, the
ratio of these two line segments is not one.