Proof: Ceva's Theorem
Proof: Ceva's Theorem
We will use the notation (XYZ) to denote the area of an arbitrary triangle XYZ.
Ceva's Theorem: In triangle ABC, if the Cevians AX, BY, and CZ are concurrent, then .
Proof: Let ABC be a triangle with Cevians AX, BY, and CZ being concurrent. Let BD, AE, and CF be the altitudes of ABC. Let OG be an altitude of AOC, OH be an altitude of COB, and OI be an altitude of BOA. Notice the following three things:
1)
2)
3)
Now by multiplying the ratios together, we see that .
Therefore, in triangle ABC, if the Cevians AX, BY, and CZ are concurrent, then .
Proof: Ceva's Theorem
Proof: Converse of Ceva's Theorem
Converse of Ceva's Theorem: In triangle ABC, if the product of , then the Cevians AX, BY, and CZ are concurrent.
Proof: Let ABC be a triangle with . We wish to show that AX, BY, and CZ are concurrent. Assume that Cevians AZ and CX intersect at O, and that the other Cevian through O is BH. By Ceva's Theorem, we have that . However, by our given we know that . Thus, we can see that . From this, we can imply that . Note that this can only be true if H and Z are the same point. Therefore, we can conclude that AX, BY, and CZ are concurrent because we assumed that AZ, CX, and BH were concurrent.