Ceva’s Theorem
A line segment connecting a vertex of a triangle to a point on the opposite side is called a
cevian. The medians of a triangle – the line segments connecting the vertices of the
triangle to the midpoints of the opposite side – are important examples of cevians. It is a
fact that the three medians of a triangle intersect at a single point, called the centroid of
the triangle.
In the diagrams below, cevians are shown. In the triangle on the right, the cevians all
intersect at a single point; in this case we say that the cevians are concurrent at that
point. In the diagram on the right, the cevians are not concurrent. Ceva’s Theorem (after
whom cevians are named) tells us the conditions under which cevians are concurrent.
nonconcurrent cevians
concurrent cevians
Open the Ceva’s Theorem Sketchpad file to see the statement of Ceva’s Theorem.

Explain, using Ceva’s Theorem, why the medians of a triangle all intersect at a
single point.

Prove the following theorem, using Ceva’s Theorem and facts about similar
triangles.
In the diagram at the right, given that EF is
parallel to BC , prove that D is the midpoint of
BC and that H is the midpoint of EF .
Proving Ceva’s Theorem
The proof of Ceva’s Theorem is done in two parts, since it is an “if and only if”
statement. In your presentation, you will need to fill in the missing details.
Part 1: If the cevians are concurrent, then the ratio equation is true
Start with a triangle ABC and construct a
line passing through A that is parallel to BC .
We know that the cevians are concurrent (at
G), so draw a line segment from B through G
until you intersect the line. Call this point B.
Construct C in a similar way.
You should notice several pairs of similar triangles in this diagram. For example, we
have CAG ~ CDG (prove it!).
Prove, using similar triangles, that each of the following equations is true:
BD DG

,
B A GA
DG DC
CE BC


,
,
GA AC 
EA B A
AF AC 

FB
BC
Combine these equations (using substitution and multiplication) to obtain the desired
equation:
AF BD CE


 1.
FB DC EA
Part 2: If the ratio equation is true, then the cevians are concurrent
Given a triangle ABC and points D, E, and F such that
AF BD CE


1
FB DC EA
is true, we must show that the cevians to D, E, and F are
concurrent. We will use the proof we have already
completed to help us. Construct the cevians BE and CF and their intersection point G.
Construct a cevian from A, through G, to BC . Call this intersection point D. We now
know that the cevians to D, E, and F are concurrent, but we don’t know how this new
point D relates to the original point D.
Prove that D and D are in fact the same point, and explain how this fact completes this
part of the theorem. (Hint: Use your equations to prove that BD  BD  .)
Further Exploration
Use Sketchpad to illustrate the following points, and use Ceva’s Theorem to explain why
they exist.


The Fermat Point of a Triangle (see p. 322 of your book)
The Orthocenter of a Triangle (see #8 on p. 326 of your book)