proof of Ceva’s theorem
As in the article on Ceva’s Theorem, we will consider directed line segments.
Let , and be points on , and , respectively, such that , and are concurrent, and let be the point where , and meet. Draw a parallel to through the point . Extend until it intersects the parallel at a point . Construct in a similar way extending .
The triangles and are similar, and so are and . Then the following equalities hold:
Notice that if directed segments are being used then and have opposite signs, and therefore when cancelled change the sign of the expression. That’s why we changed to .
Now we turn to consider the following similarities: and . From them we get the equalities
which lead to
Multiplying the last expression with (1) gives
and we conclude the proof.
To prove the converse, suppose that are points on respectively and satisfying
Let be the intersection point of with , and let be the intersection of with . Since then are concurrent, we have
which implies , and therefore are concurrent.
|Title||proof of Ceva’s theorem|
|Date of creation||2013-03-22 12:38:54|
|Last modified on||2013-03-22 12:38:54|
|Last modified by||yark (2760)|