Step  Discussion  Justification 
0 
Let ABC be a triangle. Let D be a point but not an endpoint on AB, E
be a point but not an endpoint on BC, F be a point but not an endpoint
on CA such that the segments AE, BF and CD are concurrent. Let P be the
point where AE, BF and CD are concurrent.
 These are the
criteria.

1 
Draw a line through C parallel to AB. Label this line c.

2 
Extend line segment AE. Label this line e. Label the intersection
of lines c and e as A'.

3 
Extend line segment BF. Label this line f. Label the intersection
of lines c and f as B'.

4 
Angle AEB and angle A'EC are congruent.

Angles AEB and A'EC are opposite angles.

5 
Angles A'CE and ABC are congruent

Since BC is a transversal of parallel lines AB and c, angles A'CE and ABC are congruent.

6 
Angles AA'C and A'AB are congruent.

Since BC is a transversal of parallel lines AB and c, angles A'CE and ABC are congruent.

7 
Triangles ABE and A'CE are similar triangles.

Since AEB is congruent with A'EC, A'CE is congruent with ABC and AA'C is
congruent with A'AB, triangles ABE and A'CE are similar triangles.

8 
Triangles BAF and B'CF are similar triangles.

Triangles BAF and B'CF are similar by an argument similar to steps 46.

9 
The following equalities hold:
.

The ratio of one corresponding side of a similar triangles to another corresponding side are equal. 
10 
Now multiply the respective sides of the equations in step 10 to get
.

This uses the multiplicative property of equality and the substitution property of equality.

11 
Triangle ADP is similar to triangle A'CP.

Triangle ADP is similar to triangle A'CP by an argument similar to steps 46.

12 
Triangle BDP is similar to triangle B'CP.

Triangle BDP is similar to triangle B'CP by an argument similar to steps 46.

13 
The following equalities hold:

The ratio of one corresponding side of a similar triangles to another corresponding side are equal. 
14 
This gives

This uses the transitive property of equality
and the multiplicative property of equality.

15 
Multiplying the equation from step 10 with the equation from step 14 gives
Q.E.D.

This uses the multiplicative property of equality.

Converse 
Step  Discussion  Justification 
1 
Suppose that E,F,D are points on BC,CA and AB respectively satisfying
.

These are the criteria.

2 
Let Q be the intersection of AE with BF and D' be the intersection
of CQ with AB.

3 
Since AE, BF and CD' are concurrent,
and
.


4 
Step #3 implies D = D', so AE, BF, and CD are concurrent.
Q.E.D.


Table 1: Proof of Ceva's Theorem. Proof courtesy Yark. Licensed under Creative Commons Attribution 2.5 license. 