 Ceva’s Theorem

Final Assignment

Part B

By Erin Cain

In this problem we are asked to explore Ceva’s Theorem.

Consider any triangle ABC.  Select any point P inside the triangle and draw lines AP, BP, and CP extended to their opposite sides in points D, E, and F respectively. 1)     Explore (AF)(BD)(EC) and (FB)(DC)(EA) for various triangles and various locations of P.    As you can see from the above pictures, no matter what kind of triangle you are looking at, and not matter where P is located, the products (AF)(BD)(EC) and (FB)(DC)(EA) are always equal (to manipulate this triangle on your own, click HERE).  Therefore, the ratio of the two products will always be one.  This is Ceva’s Theorem.

In order to prove that this is true, we must show that if AD, BE, and CF are concurrent if and only if the product of the ratios equals one, i.e.

(AF)/(FB) * (BD)/(DC) * (CE)/(CA) = (AF * BD * CE) / (FB * DC * CA) = 1.

To begin our proof, we will use our point of concurrency, P, to form three triangles, PAC, PCB, and PBA. Next we can construct line segments from P that are perpendicular to all three sides of our original triangle, ABC (the black, dashed lines in the picture below).  These line segments become the altitudes for these three triangles. When looking at the above picture, there are actually six different triangles inside of our original one.  Therefore, our altitudes, q, r, and s, are the altitudes for two different triangles.

q is the altitude for triangles BPD and DPC

r is the altitude for triangles BPF and FPA

s is the altitude for triangles APE and EPC

Now that we know the altitude for each of these triangles, we can use this to compare the areas of the triangles (note that the altitude of a triangle equals the height of a triangle.  Let’s look at the ratio of the area of triangle CPE to the area of the triangle EPA.

Area(CPE) = (1/2)CEs

Area(EPA) = (1/2)EAs

Area(CPE) / Area(EPA) = [(1/2)CEs] / [(1/2)EAs]

Area(CPE) / Area(EPA) = CE / EA

Due to the fact that they share the same altitude/height, both the ½ and the altitude in both areas will cancel each other out (i.e. when you divide them you will get a quotient of 1).  This holds true for the rest or the triangles.  Therefore, we have the following:

Area(BPD) / Area(DPC) = BD / DC

Area(FPA) / Area(BPF) = FA / BF

Therefore we have the following: QED.

How can this theorem be used in proving concurrency of the medians if P is the centroid?

The fact that the medians of a triangle has a point of concurrency is one of the easiest ideas to prove with the use of Ceva’s Theorem.  A median is a line that connects the midpoint of one sides of a triangle to its opposite vertex.  Therefore, a median divides the side of a triangle into a ratio of 1 to 1.  Therefore, when you a triangle has all three medians drawn and you find the product of the ratios of the sides, you will be multiplying ratios of 1 to 1, hence the product of the ratios equals one.  By Ceva’s Theorem, they must have a point of concurrency, which is the centroid of the triangle. How can this theorem be used in proving concurrency of the lines of the altitudes if P is the orthocenter?

This can be done similarly to the proof above of Ceva’s Theorem.  When you draw in the altitudes of a triangle, you essentially divide the original triangle into six smaller triangles.  Unlike the proof above, the altitudes are already drawn for each of the smaller triangles.  Therefore, you can set up the ratios and find that the product of the areas of each of the smaller triangles equals the products of the ratios of the sides of the triangle, which in turn equals 1.  Therefore there the altitudes of a triangle have a point of concurrency which is the orthocenter. How can this theorem be used in proving the concurrency of the angle bisectors if P is the incenter?

Similar to above, you can prove the concurrency of the angle bisectors of a triangle simply by proving Ceva’s Theorem when P is at the incenter.  Once again, you end up with six smaller triangles when you construct the angle bisectors.  If you take the point of intersection and construct perpendiculars to each side of the original triangle, you will have created all of the altitudes for the six triangles (the two triangles on the same base share the same altitude).  Form here you can follow the same steps for proving the concurrency of the altitudes of a triangle or the steps for proving Ceva’s Theorem. How can this theorem be used in proving the concurrency of the perpendicular bisectors if P is the circumcenter?

Similar to proving the concurrency of the medians of a triangle, perpendicular bisectors divide the sides of the triangle into a 1 to 1 ratio.  Therefore, the product of the ratios of the sides will equal 1, and for this reason, the perpendicular bisectors have a point of concurrency at the circumcenter of the triangle. 