Ceva’s Theorem

 

By Courtney Cody

 

 

In this assignment, we will investigate, prove, and extend Ceva’s Theorem.

 

Let’s begin by constructing a diagram to represent Ceva’s Theorem.  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.  We wish to explore the relationship between the products  and , or the product of the left-hand-sides (green) and the right-hand sides (orange) of each side of triangle ABC.

To begin our investigation, we will the built-in measure and calculation tools to explore the relationship between the two products.

As we can see in the diagram above, the products  and  appear to be equal.  Click HERE to move the points of the triangle around to see if the products remain the same.

 

 

THEOREM

 

Ceva’s theorem states that if points D, E, and F are on the sides AB, BC and AC of a triangle ABC then the lines AE, BF and CD are concurrent if and only if the product of the ratios

 

.

 

 

PROOF

 

We will begin by assuming the following diagram for some point P inside the triangle:

 

Suppose we create a line parallel to BC through A and extend the segments CF and BE beyond the triangle to form points G and H respectively as the lines intersect the parallel line through A.  Several sets of similar triangles form, four of which we will pay particular attention to, which are described below.

 

(1) Triangle AFG is similar to triangle CBF.

 

 

The two green triangles pictured in the diagram above are similar because opposite angles AFG and BFC are congruent, AGF and FCB are alternate interior angles, and thus GAF and FBC are congruent angles because all other corresponding angles are congruent and by the triangle sum theorem the third corresponding angles in each of the two triangles must be congruent.  Therefore, triangle AFG is similar to triangle CBF.

 

Since AFG~BFC, we can set up the following proportion:

 

 

(2) Triangle BDP is similar to triangle HAP.

 

By a similar method as in (1), triangle BDP is similar to triangle HAP.  Since BDP~HAP, we can set up the following proportion:

 

 

 

(3) Triangle AHE is similar to triangle CBE.

 

By a similar method as in (1), triangle AHE is similar to triangle CBE.  Since AHE~CBE, we can set up the following proportion:

 

 

 

(4) Triangle APG is similar to triangle DPC.

 

By a similar method as in (1), triangle APG is similar to triangle DPC.  Since APG~DPC, we can set up the following proportion:

 

Hence, after exploring the different similar triangles, we have the following four proportions:

(a)            (b)         (c)          (d)

 

Since the ratio  is common to both (b) and (d), then we know that .

 

Since G is not a part of the original triangle, we want to eliminate AG from the above proportion.  Hence, we can use (a) to solve for AG and substitute.

Since , then .

 

Similarly, since H is not a part of the original triangle, we want to eliminate AH from the above proportion.  Hence, we can use (c) to solve for AH and substitute.

 

Since , then .

 

Now, by the reflexive property CB=BC, DC=CD, DB=BD, and BF=FB.

 

Hence, making these substitutions and rearranging the terms, we have

 

 

Thus, .

 

 

PROOF OF THE CONVERSE OF CEVA’S THEOREM

 

Consider any triangle ABC and let D, E, and F lie on segments BC, AC, and AB respectively, all of which satisfy

 

.

 

Let P be the intersection inside triangle ABC of the segments CF and BE.  Also, let D’ be the intersection of AP and BC.

 

Since AD, BE and CF are concurrent, then we know from our assumption that

 

 

However, we assumed

. 

 

Equating these two equations, we obtain:

 

 

 

Thus, D=D’, which implies that AD, BE and CF are concurrent.  Therefore, Ceva’s theorem holds.

 

 

CONCURRENCY OF THREE MEDIANS

 

Suppose P is the centroid of triangle ABC. 

 

 

How could we use Ceva’s theorem to prove concurrency of the medians?  If we recall the definition of a centroid of a triangle, it is the point of intersection of the triangle’s three medians.  In relating this definition to Ceva’s theorem, this means that points D, E, and F are the midpoints of segments BC, AC, and AB respectively.  Hence, AF=FB, BD=CD, and CE=AE.  Now, Ceva’s theorem states that segments AD, BE and CF are concurrent if and only if the product of the ratios

 

.

 

In substituting the values of the aforementioned equivalent segments, we have

 

 

 

Therefore, since the ratio of the products is one, then the three medians, namely AD, BE and CF, are concurrent.

 

 

GENERALIZATION

 

Suppose we want to explore a generalization of Ceva’s theorem such that P can lie outside of triangle ABC.  In order for P to lie outside the triangle, we must construct our diagram using lines rather than segments.  Click HERE to see a working GSP sketch of this case.

 

 

Therefore, we see from the diagram and the GSP sketch that even when P is not contained within triangle ABC, Ceva’s theorem still holds since the ratio of the products is still one.


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