

Ceva's Theorem by Margaret Morgan 

triangle ABE is similar to IPE (because IP is parallel to AB) So, IP/AB = PE/BE  (5) triangle ABD is similar to triangle LDP So, LP/AB = PD/AD   (7) Substituting (1) into 6, we get AF/FB = IP/PL = (PE/BE)/(PD/AD) AF/FB = (PE/BE)/(PD/AD) = ((PE)(AD))/((BE)(PD))  (8) triangle FBC is similar to triangle FGP ( because GP is parallel to BC) So, GP/BC = PF/FC  (9) triangle BEC is similar to triangle PEJ (because PJ is parallel to BC). So, PJ/BC = PE/BE  (10) Dividing (8) and (9), we get (GP/BC)/(PJ/BC) = (PF/FC)/(PE/BE) GP/PJ = ((PF)(BE))/((FC)(PE))  (11) Substituting (2) into (10) we get BD/DC = ((PF)(BE)/(PE)(CF))  (12) triangle ADC is similar to triangle PDK (because PK is parallel to EC). So PK/ AC= PD/AD
 (13) triangle AFC is similar to FHP (because HP is parallel to AC) HP/ AC= PF/CF  (14) Dividing (12) and (13) we get PK/HP = (PD/AD)/(PF/CF) = ((PD)(CF)/(PF)(AD))  (15) Substituting (3) into (14) we get CE/EA = ((PD)(CF)/(PF)(AD))  Multiplying (7), (11), and (15) we get (AF/FB)(BD/DC)(CE/EA) = ((PE)(AD) (PF)(BE)(PD) (CF)) /( (BE)(PD)(PE) (CF) (PF)(AD)) = 1 So finally, we have (AF)(BD)(CE) = (FB)(DC)(EA) Ceva's
Theorem states in a triangle ABC, three lines AD, BE and CF intersect
at a single point P if and only if Proof: Argument above shows that if AD, BE, and CF intersect at a point P then the given ratio is 1. Now, we will prove it in the other direction. Start with a triangle ABC, with points DEF such that (AF)(BD)(CE)/(FB)(DC)(EA) = 1.
Let P be the intersection of BE and FC. Draw a line From A through P until it intersects BC at point H. Now we know, from what we just proved that (AF)(BH)(CE)/(FB)(HC)(EA) = 1 So, (AF)(BD)(CE)/(FB)(DC)(EA) = (AF)(BH)(CE)/(FB)(HC)(EA) Simplifying we get BD/DC = BH/HC Now to compare the two ratios let's make both numerators the full length of BC. BD/DC + DC/DC = BH/HC + HC/HC This is equivalent to adding one to both sides, so we still have equality. BC/DC = BC/HC This can only be true if D and H are the same point.
Special Case of P: What if D,E, and F are the midpoints of the sides of ABC?
Initially, we considered only points P that lie inside the triangle ABC, but ceva's theorem holds for points P outside of triangle ABC as well. Click hear to see a GSP sketch that illustrates the theorem for any possible point P. Seeing that the theorem holds for P outside of ABC allows us to look at another special case of P. What if D, E, and F are the feet of the altitudes of triangle ABC? (1) Triangle ACD is similar to triangle BCE <BCE = <ACD  same angle <BEC = <ADC  both right angles by construction So, <EBC = <DAC BC/AC = BE/AD = EC/DC  (2) Triangle AFC is similar to triangle EAB <FAC = <EAB  same angle <AFC = BEA  both right angles by construction So, <ACF = < EBA AF/EA = CF/BE = AC/AB  (3) Triangle BAD is similar to triangle FBC <ABD = <FBC  vertical angles <ADB = <BFC  both right angles by construction So, <BAD = <BCF AD/FC = AB/BC = BD/BF  Multiplying (1), (2), and (3) AF/EA * BD/BF *EC/DC= AC/AB * AB/BC *BC/AC (AF * BD * EC)/ (EA*BF *DC) = 1 Now, by Ceva's theorems the lines are concurrent. We refer to the point p as the orthocenter.
