Question

Prove the necessary part of Ceva’s Theorem for the case where D and E are ideal but F is ordinary.

The case is where:  D and E are ideal but F is ordinary.

Lately I feel as though the people working for Chegg prematurly take on questions they cannot answer. If you do not know the answer that is ok but please do not accept the question if you do not think you can answer it. Someone anonymously accepted this question and then said the description of the case is needed.... It is frustrating to get this response when it is clearly stated in the prompt: for the case where D and E are ideal but F is ordinary. I feel that the only reason they would respond like this is if they have no idea what the question is asking. This is becoming a regular occurence on this site so I feel someone should start monitoring the questions being asked by their so called "expert" tutor. FYI I already soved this problem I just think that this should be brought to Chegg's attention.

Answer 1

Let be a triangle, and let be points on lines , respectively. Lines are concurrent if and only if

,

where lengths are directed. This also works for the reciprocal of each of the ratios, as the reciprocal of is .

PROOF:

We will use the notation to denote the area of a triangle with vertices .

First, suppose meet at a point . We note that triangles have the same altitude to line , but bases and . It follows that . The same is true for triangles , so

.

Similarly, and , so

.

Now, suppose satisfy Ceva's criterion, and suppose intersect at . Suppose the line intersects line at . We have proven that must satisfy Ceva's criterion. This means that

,

so

,

and line concurrs with and .