This theorem gives the condition under which three lines passing through the three vertices of a triangle and points on the opposite sides (or their extensions) intersect at one point.
In △ABC, let points D, E, and F be taken on lines BC, CA, and AB, respectively, different from the vertices.
Lines AD, BE, and CF intersect at one point (or are all parallel)
⟺ BD/DC × CE/EA × AF/FB = 1
Example ①
In △ABC, if BD:DC=2:1 and CE:EA=3:1, find AF:FB assuming that lines AD, BE, and CF intersect at one point.
By Ceva's theorem: (2/1)×(3/1)×(AF/FB) = 1
6×(AF/FB) = 1 → AF/FB = 1/6
Solution:AF:FB = 1:6
In △ABC, if BD:DC=2:1 and CE:EA=3:1, find AF:FB assuming that lines AD, BE, and CF intersect at one point.
By Ceva's theorem: (2/1)×(3/1)×(AF/FB) = 1
6×(AF/FB) = 1 → AF/FB = 1/6
Solution:AF:FB = 1:6
Example ② (Application to the Centroid)
For the centroid G of △ABC, calculate BD/DC × CE/EA × AF/FB when D, E, and F are the midpoints of the respective sides.
BD/DC=1、CE/EA=1、AF/FB=1 → 1×1×1 = 1 (satisfying Ceva's theorem)
For the centroid G of △ABC, calculate BD/DC × CE/EA × AF/FB when D, E, and F are the midpoints of the respective sides.
BD/DC=1、CE/EA=1、AF/FB=1 → 1×1×1 = 1 (satisfying Ceva's theorem)
Key Points
- Ceva's theorem: BD/DC × CE/EA × AF/FB = 1 ⟺ the three lines intersect at one point
- Starting from a vertex, read the ratios alternately in the order “clockwise → counterclockwise”
- The centroid, incenter, and orthocenter all satisfy Ceva's theorem
Practice Problems
- In △ABC, if BD:DC=1:2 and AF:FB=3:1, find CE:EA assuming that lines AD, BE, and CF intersect at one point.
- For the incenter I of △ABC, let D be the point where the extension of AI intersects BC. Find BD:DC when ∠B=60° and ∠C=80° (use the angle-bisector theorem).
- In △ABC, D is the midpoint of BC, and E divides CA internally in the ratio 2:1. If the intersection of AD and BE is P, find AF:FB for the point F where the extension of CP intersects AB.
Solutions and Explanations
- (1/2)×(CE/EA)×(3/1)=1 → CE/EA=2/3 → CE:EA=2:3
- By the angle-bisector theorem, BD:DC=AB:AC=sin80°:sin60° (via the sine rule). Alternatively, BD:DC=c:b.
Since ∠A=40°, a:b:c=sin40°:sin60°:sin80°. BD:DC=sin80°:sin60°≒0.985:0.866≒approximately 34:30 (simplified value) - By Ceva's theorem, (BD/DC)×(CE/EA)×(AF/FB)=1. D: midpoint → BD/DC=1. E: CE/EA=1/2 (E divides CA from the C side in the ratio 2:1).
1×(1/2)×(AF/FB)=1 → AF/FB=2 → AF:FB=2:1