FINAL PROJECT
Ceva’s
Theorem
By
MICHELLE
NICHOLS
Consider
any triangle, ABC. Select a point P
inside the triangle and draw lines AP, BP, and CP extended to their
intersections with the opposite sides in point D, E. and F respectively.
Explore
(AF)(BD)(EC) and (FB)(DC)(EA) for various triangles and various locations of P.
No
matter where you move the P, as long as it stays inside Triangle ABC, (AF)(BD)(EC)
and (FB)(DC)(EA) remain equal. By Ceva’s
Theorem, this is always true.
CEVA’S THEOREM states that if the points D,E, and F are on the sides of
AB, BC, and AC of a triangle, then the lines AD, BE, and CF are concurrent if
and only if the product of the ratios
How
can we prove this theorem holds true?
Start with constructing parallel lines, and looking at triangles and
their measurements.
Construct
lines parallel to AD through points B and C.
By observing the sketch below, we can make several conjectures.
Using
alternate interior angles and vertical angles we will be able to prove the
ratio above is equal to 1. According to
our construction:
·
▲BMF is similar to ▲AFP, so 
·
▲CEN is similar to ▲AEP, so 
Because
of common angles and corresponding lines:
·
▲BPD is similar to ▲BNC & ▲MBD is similar to
▲PDC
So,
when we multiply these figures, and simplify, we end up with:
Thus,
proving Ceva’s Theorem…
What about if P is outside the triangle? Look below at the triangle with P outside, or
click here to go to GSP to observe on your
own.
Show that when P is inside triangle ABC, the ratio
of the areas of triangle ABC and triangle DEF is always greater than or equal
to 4. When is it equal to 4?
This
holds true only when P is INSIDE the triangle.
When
will the area be exactly 4? When
Triangle DEF is the medial triangle of Triangle ABC. See the sketch below, or click here to go to the GSP file.
