Ceva’s Theorem

 

By Pei-Chun Shih

 

 

Consider any triangle ABC. Select a point P inside the triangle and draw lines AP, BP, and CP extended to their intersections with the opposite sides in points D, E, and F respectively.

 

Ceva’s Theorem says that in a triangle ABC, three lines AD, BE, and CF are concurrent at a single point P if and only if .

 

 


 

First, let’s explore (AF)(BD)(EC) and (FB)(DC)(EA) for various triangles and various locations of P inside triangle ABC.





 

 

 

After exploring various triangles and various locations of P via GSP, I found that the products of (AF)(BD)(EC) and (FB)(DC)(EA) are always the same no matter what kind of triangle it is and no matter where the location of P is as long as P is inside triangle ABC.

 

Next, I moved point P outside triangle ABC and found that the products of (AF)(BD)(EC) and (FB)(DC)(EA) are still the same with various triangles.

 

 

 

 


 

Here I am going to prove the relationship discovered in the previous section by considering areas of triangles inside triangle ABC.

 

 


 

After proving the Ceva’s Theorem, it is time to apply it to the concurrency problems we have explored in the previous write-ups. Recall that in the write-up 04, we proved the altitudes of a triangle are concurrent by constructing circles and using the arc angle theorem. This proof would be much easier by applying the Ceva’s theorem.

 

 

The diagram above shows triangle ABC with its orthocenter H. Therefore, AD, BE, and CF are altitudes of triangle ABC. The same technique we use to prove the Ceva’s theorem can be used here to prove that the three altitudes are concurrent. Our goal is to find that the ratio of (AF)(BD)(EC) and (FB)(DC)(EA) is equal to 1.

 

 

Proving the concurrency of three medians of a triangle is a breeze if we use the Ceva’s Theorem. Below is the diagram of the centroid of a triangle ABC. The points D, E, and F are the midpoints of triangle ABC since AD, BE, and CF are medians. Therefore, BD = DC, EA = EC, and AF = FB. The ratio of (AF)(BD)(EC) and (FB)(DC)(EA) is equal to 1 since (AF)(BD)(EC) = (FB)(DC)(EA). Thus, the three medians of triangle ABC are concurrent at G (the centroid) by the Ceva’s theorem.

 

 

The proofs of the concurrency of the three angle bisectors (if P is the incenter) or the three perpendicular bisectors (if P is the circumcenter) of a triangle ABC are similar to the proof of the orthocenter we have done earlier. Therefore, I would like to skip them here.

 

Basically the Ceva’s theorem is widely used in geometry to prove the concurrency of lines in a triangle.

 

 

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