CEVA’S THEOREM

BY

SHADRECK S CHITSONGA

Ceva’s Theorem states that if we let ABC be a triangle in the Euclidean plane, and let F be a point on AB other than A and B, let D a point on BC other than B and C, and let E be a point on CA other than C and A. If we assume that AD,CF, BE are not all parallel then these lines have a point in common if and only if the equation holds.

We are going to consider a number of cases by varying the position of P and using different triangles.

Case 1

Acute angled-triangle with P as the circumcentre.

In this particular case we see that the products (AF)(BD)(CE) and (BF)(CD)(AE) are equal or in other words

CLICK HERE to see different sizes of triangle ABC and compare the two products (AF)(BD)(CE) and (BF)(CD)(AE). Are they always equal?

Case 2

The point P is now the centroid of the triangle ABC

Just is we found in case one we still notice that when P is now the centroid ,

still holds.

OPEN HERE to manipulate the triangle. Look at different triangles I.e., obtuse, right-angled triangle to see whether , still holds. You can drag any of the points A, B, or C to alter the shape of the triangle.

Case 3

The point P is the incenter of triangle ABC

The triangle considered here is a right-angled triangle and P is the incenter. We see that in this case just as in cases 1 and 2, , still holds.

CLICK HERE to explore different triangles i.e obtuse,

Case 4

The point P is now the orthocenter of the triangle ABC.

Case 4 is not any different from the other three cases considered already, the ratio , still holds.

Case 5

Now we will choose P as any arbitrary point. This time P is a point different from all the other cases already.

Case 5, still shows that the ratio we have be considering in all the other four cases still holds even if we choose P as an arbitrary point.

NOTE: Here we have just used GSP to show that the ratio does not change. This should not be regarded as proof. Use this link to see proofs for the different positions of P that we have considered.

Case 6

In all the cases above we let the point P be inside the triangle. Now let us consider the case when the point P is outside the triangle. Will the ratio still be equal to 1?

Even when P is outside the theorem still holds.

CLICK HERE to change the positions of P.

For some application of the theorem go to this link.

END