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…
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.