Final Part 2


Ceva's Theorem

 

Three cevians AF, BD, and CE of triangle ABC are concurrent if and only if

 

(BF)(CD)(AE) / (FC)(DA)(EB) = 1

 

 

 

 

Proof that, if the three cevians are concurrent, then

(BF)(CD)(AE) / (FC)(DA)(EB) = 1

 

Through A and C, draw lines parallel to BD. Let the points in which they intersect CE and AF be called K and J.

 

 

 

Because of equal vertical angles and equal alternate interior angles Triangle BFP is similar to Triangle CFP and Triangle AEJ is similar to Triangle BEP. Therefore,

(BF)/(FC) = (BP)/(CK)

and

(AE)/(BE) = (AJ)/(BP)

 

Because of the common angle and equal corresponding angles, Triangle CDP is similar to Triangle CAJ and Triangle CAK is similar to Triangle DAP. Therefore,

(CD)/(CA) = (DP)/(AJ)

and

(CA)/(DA) = (CK)/(DP)

 

We can now build the conclusion of Ceva's theorem by multiplying the left and right sides of these equations.

(BF/CF) * (CD/CA) * (CA/DA) * (AE/BE) = (BP/CK) * (DP/AJ) * (CK/DP) * (AJ/BP)

 

Simplifying we get:

(BF/CF) * (CD/DA) * (AE/BE) = 1

 


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