Ceva's Theorem :

If three line segments are drawn from the vertices of a triangle passing through the same point, then these segments opposite sides in a definite ratio.

Machine Input:

Point Order : ABCPDEF
Hypotheses : coll(AFB) coll(CPF) coll(BDC) coll(APD) coll(CEA) coll(BPE)
Conjecture : AF/FB BD/DC CE/EA=1

Machine output :

We can equivalently prove that :-

1.0(EC/EA)
-------------------------- = 1
1.0(FB/FA)(DC/DB)

Proof:

   1.0(EC/EA)
   ---------------------
   1.0(FB/FA)(DC/DB)

By putting (EC/EA) = -[PCB]/[PBA]

1.0[PCB]
= ---------------------
1.0(FB/FA)(DC/DB)[PBA]

By putting (FB/FA) = [PCB]/[PCA]

1.0[PCA]
= ---------------------
1.0(DC/DB)[PBA]

By putting (DC/DB) = [PCA]/[PBA]

1.0
= ---------------------
1.0

Compute Time=0.158 seconds