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. |
4/14/2019: Updated equations and expressions to new format. (
12/21/2018: Reviewed and corrected IPA pronunication. (
6/25/2018: Removed broken links, updated license, implemented new markup, updated GeoGebra apps. (
1/9/2010: Added "References". (
11/21/2008: Initial version. (