| 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 4-6.
         | 
        | 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 4-6.
         | 
        | 12 | 
         Triangle BDP is similar to triangle B'CP.
         | 
         Triangle BDP is similar to triangle B'CP by an
		 argument similar to steps 4-6.
         | 
        | 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, and 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. | 
	   
        
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