A necessary and sufficient condition for where and are points of the respective side lines of a triangle , to be concurrent is that
where all segments in the formula are directed segments.
Let be points on respectively such that 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.
Suppose AB, AC, and BC have lengths 13, 14, and 15. If and . Find BD and DC.
If and , then , and . From this, we find and .