Summer Tuggle
Final
Assignment part II
Ceva’s Theorem
In triangle ADG, AF, DI, and GC are concurrent at point E and the
following ratio is true: 
= 1.
This means that the product of the “left” segments divided by the
product of the “right” segments equals 1 when rays extended from each vertex
meet at a common point on the interior of the triangle (point E).
In
order to begin, we will extend the sides of the triangle and also the rays from
each vertex of the triangle. A line will
be drawn through point A parallel to the opposite side GD. The extended rays
from points G and D will intersect this new line at points B and H
respectively.
From these new constructions similar triangles can be found
Angle ACB and
angle DCG are vertical angles and therefore congruent. Angle BAC and angle CDG are alternate
interior angles and also congruent.
Since 2 of the three angles in these triangles are congruent then the third
pair also must be congruent. Therefore ΔABC is similar to ΔDGC.
Following the same
ideas ΔDIG is similar to ΔHIA (vertical angles HIA and DIG and
alternate interior angles HAI and DGI)
Also ΔDEF and ΔHEA are similar for the same reasons.
Likewise,
ΔFGE and ΔAEB are similar.
From
these similar triangles:
ΔFGE and ΔAEB
ΔDEF and ΔHEA
ΔDIG and ΔHIA
ΔABC and ΔDGC
Many helpful
ratios can be created.
= 
= 
= 
=
Based on these ratios, = which can be manipulated to give 
=
We can now
substitute back into the original equation:
= 1
= 1
= 1
1=1
Click here
for a Geometer’s Sketchpad file that you can manipulate.