Geometry path · G10
Before this lesson: Parallel lines and angle bisectors, Triangle area ratios, How to write a proof
Your goal: Distinguish concurrent lines from collinear points and select a suitable criterion.
Start with the idea and one example. Take a break before the written task if you need it. The questions are original teaching exercises, not official past-paper questions.
Names you may know: Ceva’s theorem; Menelaus’ theorem; centroid and orthocentre.
Ceva, Menelaus and Triangle Concurrency: the key idea
Concurrency means three or more lines meet at one point; collinearity means points lie on one line. In triangle ABC with D∈BC,E∈CA,F∈AB, Ceva states AD,BE,CF are concurrent exactly when (BD/DC)(CE/EA)(AF/FB)=1 for internal side points. Menelaus states a transversal meets the three side lines in D,E,F with the product of absolute ratios equal to 1, provided the intersection pattern has an odd number of external points; directed ratios encode the signs more generally. Do not confuse the two incidence conclusions. Medians, perpendicular bisectors and internal angle bisectors meet at the centroid, circumcentre and incentre respectively, with different defining properties.
A worked example
If BD:DC=2:3 and CE:EA=3:4, find AF:FB for concurrence of the three cevians.
Ceva gives (2/3)(3/4)(AF/FB)=1. The first product is 1/2, so AF/FB=2.
Your turn: change one thing
Why do the three medians concur by Ceva?
Try this on paper before opening the explanation.
Compare your reasoning
Each side is split in ratio 1:1. Their product is 1, so Ceva’s converse gives concurrence. The intersection is the centroid.
Methods and connections
Centres and what they guarantee
The centroid lies on the three medians and divides each in the ratio 2:1 measured from the vertex. The circumcentre lies on perpendicular bisectors and is equidistant from the vertices; it may lie outside an obtuse triangle. The incentre lies on internal angle bisectors and has equal perpendicular distances to the three sides. The orthocentre lies on the three altitude lines. Do not transfer a property of one centre to another.
Carnot’s perpendicular-line criterion
Take D on BC, E on CA and F on AB of a nondegenerate triangle. The perpendiculars to those sides at D,E,F concur exactly when BD²−DC²+CE²−EA²+AF²−FB²=0. To see this, let P be the intersection of the first two perpendiculars. Pythagoras gives PB²−PC²=BD²−DC² and PC²−PA²=CE²−EA². Adding shows the displayed condition is equivalent to PA²−PB²=AF²−FB². The locus of points satisfying this last difference is the perpendicular to AB through F, so the third line contains P.
A Menelaus example
Suppose D∈BC and E∈CA with BD/DC=2 and CE/EA=3. For D,E,F to be collinear, F must be on the extension of AB and |AF|/|FB|=1/6. The internal/external pattern is part of the theorem: a line crossing a triangle twice meets the third side line outside the segment. A positive product alone does not encode that placement.
Pappus: a further collinearity tool
Let A,B,C lie on one line and A′,B′,C′ on another. When the three cross-intersections X=AB′∩A′B, Y=AC′∩A′C and Z=BC′∩B′C are defined and distinct, Pappus’s theorem says X,Y,Z are collinear. This is an advanced theorem stated for recognition here, not derived from Ceva. First mark the two triples and identify all six joining lines before trying to apply it.
Pause and check
A trap to avoid: Confusing Ceva with Menelaus or mixing directed and unsigned conventions.
Practise and adjust the level
Foundation checks the language; Core applies the method; Stretch asks you to choose or justify an idea. These are levels within this lesson. A session selects six of the nine questions; unused questions allow the level to change. Advanced theory still needs written practice.
Interactive practice loads here. You can also use the complete question set below.
Write a complete argument
Prove the necessary direction of Ceva using area ratios for internal concurrence.
Planning hint
List the assumptions and the exact conclusion. Identify the definition or theorem in this lesson that connects them. Explain why its conditions hold before using it.
Read the full solution after your attempt
Let the cevians meet at P inside ABC. Triangles PBD and PCD have area ratio BD/DC. Triangles PBA and PCA have the same ratio: subtracting the proportional small triangles from ABD and ACD, or comparing the altitude from P to BC with that from A, gives [PBA]/[PCA]=BD/DC. Similarly CE/EA=[PCB]/[PAB] and AF/FB=[PAC]/[PBC]. Multiplying cancels all three nonzero areas and yields 1.
My proof notebook
Write on paper, or keep a draft here. Compare your reasoning with the solution only after a real attempt. The checklist is your own review, not an automatic mark.
All nine practice questions
Prefer paper or have JavaScript switched off? The complete question set, hints and solutions are here. Interactive practice uses these same questions in an order chosen from your answers.
1. Concurrent lines do what?
Foundation
- Meet at one point
- Are all parallel
- Lie on one circle
- Have equal lengths
Hint
Concurrency is an incidence condition on lines.
Answer and reasoning
Meet at one point. The same point belongs to all the lines.
2. Ceva’s theorem detects what?
Foundation
- Equal areas always
- A right angle
- Concurrence of cevians
- Collinearity of all triangle vertices
Hint
Its lines join vertices to opposite sides.
Answer and reasoning
Concurrence of cevians. The product condition determines whether those three lines meet.
3. For ratios 2/3 and 3/4, what third ratio makes Ceva’s product 1?
Core
- 1/2
- 3
- 4
- 2
Hint
The first two multiply to 1/2.
Answer and reasoning
2. The remaining ratio must be 2.
4. Three medians have a side-ratio product of what?
Core
- 1
- 0
- 2
- 3
Hint
Each midpoint ratio is 1.
Answer and reasoning
1. 1·1·1=1.
5. What is the usual conclusion of Menelaus?
Stretch
- All sides are equal
- The triangle is cyclic
- Three points on the side lines are collinear
- Three cevians concur
Hint
Distinguish it from Ceva.
Answer and reasoning
Three points on the side lines are collinear. Menelaus characterises a straight transversal through the side lines.
6. Why must external positions or directed signs be tracked?
Stretch
- Lengths cannot be positive
- Ceva never allows internal points
- Signs are purely decorative
- Absolute ratio products alone can hide the incidence pattern
Hint
Different configurations can have the same positive ratios.
Answer and reasoning
Absolute ratio products alone can hide the incidence pattern. The location of points determines whether a signed ratio criterion applies.
7. Collinear points lie where?
Foundation
- At equal distances always
- Inside one triangle only
- On one line
- On one circle only
Hint
Collinearity is a straight-line condition.
Answer and reasoning
On one line. One line contains all the specified points.
8. Ceva’s first two ratios are 1/2 and 1/3. The third must be what?
Core
- 6
- 1/6
- 2/3
- 3/2
Hint
The three ratios multiply to 1.
Answer and reasoning
6. (1/2)(1/3)r=1 gives r=6.
9. Why is the centroid not generally the circumcentre?
Stretch
- A triangle has only one centre
- Medians are always perpendicular bisectors
- All centres lie outside
- Their defining lines and equalities differ
Hint
Compare median concurrency with equal vertex distance.
Answer and reasoning
Their defining lines and equalities differ. They coincide in an equilateral triangle but are generally distinct.
Choose your next step
Continue to Circles, tangents and power of a point. If this felt difficult, return to a prerequisite above. Every lesson stays open.
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