Ceva's TheoremDate: 03/04/99 at 21:41:16 From: Mark Ride Subject: Ceva's Theorem Hi, I am a grade 12 student and I can't seem to get a good solution for the following questions: (a) Ceva's Theorem" The three lines drawn from the vertices A, B, and C of triangle ABC, meeting the opposite sides in points D, E, and F respectively, are concurrent if and only if AF/FB*BD/DC*CD/EA = 1. This theorem is credited to seventeenth-century Itailian mathematician Giovanni Ceva. Prove it using vector methods. (b) The importance of Ceva's Theorem lies in its use to prove many classic results in geometry. Use Ceva's Theorem to prove each of the following results: (i) The medians of any triangle are concurrent. (ii) The altitudes of any triangle are concurrent. (iii) The interior angle bisectors of a triangle are concurrent. Date: 05/09/99 at 08:36:38 From: Doctor Floor Subject: Re: Ceva's Theorem Hi Mark, Thanks for your question. Let's consider Ceva's theorem: This theorem is about a triangle ABC, and points A' on sideline BC, B' on AC and C' on AB. It states that AA', BB' and CC' intersect in one point T if and only if: AC' BA' CB' --- * --- * --- = 1 C'B A'C B'A We can easily prove this when A', B' and C' are on segments BC, AC, and BC respectively: First, let AA', BB', and CC' intersect in one point T. We see that area(AC'C):area(C'BC) = area(AC'T):area(C'BT) = AC'/C'B. But then, since area(ATC) = area(AC'C)-area(AC'T) and area(BTC) = area(C'BC)- area(C'BT), we find that AC'/C'B = area(ATC)/area(BTC). In the same way we can find BA'/A'C = area(ATB)/area(ATC) and CB'/B'A = area(BTC)/ area(ATB). And we find AC' BA' CB' area(ATC) area(ATB) area(BTC) --- * --- * --- = --------- * --------- * --------- = 1 C'B A'C B'A area(BTC) area(ATC) area(ATB) For the converse, suppose that we know that the following equation holds: AC' BA' CB' --- * --- * --- = 1 C'B A'C B'A Then, in the same way as we did above, this means that for each point P on CC', and only for points on CC', we know that area(APC)/area(BPC) = AC'/C'B. Also AA' exists of the points Q satisfying area(AQB)/area (AQC) = BA'/A'C and BB' exists of points R satisfying area(BRC)/area (ARB) = CB'/B'A. Now, let T be the point of intersection of CC' and AA'. Then we have: area(ATC) area(ATB) CB' --------- * --------- * --- = 1 area(BTC) area(ATC) B'A So CB'/B'A = area(BTC)/area(ATB), and T lies on BB' too. So we have proven Ceva's theorem, except when A', B', C' and/or T lie outside ABC. A proof of this is essentially the same, but then we have to take care about signs: * When T lies on the opposite side of AB from C, then area(ATB) is negative. * When A' is outside segment BC, then BA'/A'C is negative. Ceva's theorem is used to prove the existence of many triangle centers. The concurrence of the medians in the centroid is very easy to prove, because then: AC' BA' CB' 0.5*AB 0.5*BC 0.5*AC --- * --- * --- = ------ * ------ * ------ = 1*1*1 = 1 C'B A'C B'A 0.5*AB 0.5*BC 0.5*AC If you haven't done so already, you should try to do the concurrence of the altitudes and the angle bisectors yourself. Two other examples of the use of Ceva's theorem can be found in the Dr. Math archives: http://mathforum.org/dr.math/problems/schultess9.4.98.html http://mathforum.org/dr.math/problems/wanwipa7.20.98.html Best regards, - Doctor Floor, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/