This theorem gives the ratios of segments when a line not lying on one side of a triangle intersects the three sides (or their extensions).
When points D, E, and F lie on sides BC, CA, and AB (or their extensions), respectively, of △ABC, and these points are collinear:
BD/DC × CE/EA × AF/FB = 1
(However, the ratios are directed ratios, or the theorem can be used without considering signs when exactly one point lies outside the sides.)
Example ①
Take D as the point that divides side BC internally in the ratio 2:1, and take E on the extension of side CA so that CE:EA=3:1.
Find AF:FB for point F where line DE intersects the extension of side AB.
By Menelaus's theorem: BD/DC × CE/EA × AF/FB = 1
（2/1）×（3/1）×（AF/FB）= 1 → AF/FB = 1/6
Solution:AF:FB = 1:6 (F lies outside beyond B)
Take D as the point that divides side BC internally in the ratio 2:1, and take E on the extension of side CA so that CE:EA=3:1.
Find AF:FB for point F where line DE intersects the extension of side AB.
By Menelaus's theorem: BD/DC × CE/EA × AF/FB = 1
（2/1）×（3/1）×（AF/FB）= 1 → AF/FB = 1/6
Solution:AF:FB = 1:6 (F lies outside beyond B)
Example ② (Ratio of Triangle Sides)
In △ABC, BD:DC=1:2, and line EF (E lies on CA and F lies on the extension of AB) satisfies Menelaus's condition. If CE:EA=4:1, find AF:FB.
(1/2)×(4/1)×(AF/FB)=1 → AF/FB=1/2
Solution:AF:FB = 1:2
In △ABC, BD:DC=1:2, and line EF (E lies on CA and F lies on the extension of AB) satisfies Menelaus's condition. If CE:EA=4:1, find AF:FB.
(1/2)×(4/1)×(AF/FB)=1 → AF/FB=1/2
Solution:AF:FB = 1:2
Difference Between Ceva's and Menelaus's Theorems
- Ceva: condition for three lines to intersect at one point in △ → product of ratios=1
- Menelaus: condition for three points to lie on one straight line → product of ratios=1
- The forms of the equations are the same; the conditions differ (a point versus a line)
- When external division is involved, note that an odd number of points lie outside the sides
Practice Problems
- In △ABC, if BD:DC=3:2 and CE:EA=1:2, find AF:FB assuming D, E, and F are collinear.
- Take point F on side AB such that AF:FB=1:3. If line CF intersects the line through A parallel to BC at G, find AG:CF (Menelaus's theorem is not necessary).
- In △ABC, D is the midpoint of BC. For point E where line AD intersects the line through B parallel to AC, find BD:BE using Menelaus's theorem.
Solutions and Explanations
- (3/2)×(1/2)×(AF/FB)=1 → AF/FB=4/3 → AF:FB=4:3
- Consider the similarity ratio. From △AFC∽△GFA (AA), AG/CF=AF/FB=1/3 → AG:CF=1:4 (AG is 1/4 of CF)
- Since D is the midpoint of BC, BD:DC=1:1. Applying Menelaus to △ABE and considering line DC (extended), proceed in the form (BD/DE)×(EC/CA)×(AF/FB). Alternative: AD is a median, so by the ratio from the parallel line BD:BE=1:2