The Fermat-Torricelli Point of a Triangle

"Perhaps, though, there is another purpose to proof — as a testing ground for the stamina and ingenuity of the mathematician.
We admire the conqueror of Everest, not because the top of Everest is a place we want to be, but just because it is so hard to get there."
— Davis and Hersh, 1983

The dynamic geometry activities below are from the "Proof as Challenge" section of my book Rethinking Proof (free to download).

Worksheet & Teacher Notes
Open (and/or download) a guided worksheet and teacher notes to use together with the dynamic sketch below at: The Fermat-Torricelli Point Worksheet & Teacher Notes.

Prerequisites
Knowledge of the properties of cyclic quadrilaterals (e.g. a convex quadrilateral is cyclic if and only if a pair of its opposite angles are supplementary). Also, students should be familiar with the SAS method of proving a pair of triangles congruent. Specifically, it is recommended that students have already completed the following prerequisite online activities:
Concurrency Conjecture
Cyclic Quadrilateral
Cyclic Quadrilateral Converse

The Fermat-Torricelli Point of a Triangle

Notes
1) Users are strongly encouraged to download the accompanying Worksheet, and to print it out to use in conjunction with the dynamic sketch above. Alternatively, copy the URL: http://dynamicmathematicslearning.com/fermat-torricelli-point-rethinking-proof.pdf and paste it into a new browser window. Then resize the new window to place it side by side with this one, or one below the other. (However, this is not likely to be a feasible option for users using small screens such as a cellphone or tablet.)
2) Construct lines DC, EA, and FB by using the Line Tool on the left.
3) What do you notice? Check your observation by dragging.
4) Measure the distances DC, EA, and FB by using the Distance Tool on the left. (Scroll down if necessary).
5) What do you notice about these distances? Carefully check your observations by further dragging.
6) Drag C past B. What happens to the triangles?
7) When you drag C past B, do your observations from Questions 3 and 5 still hold?

Conjectures
8) You should have noticed that if equilateral triangles DBA, ECB, and FAC are constructed on the sides of any right triangle ABC, then
.....i) DC, EA, and FB are concurrent.
.....ii) DC, EA, and FB are equal in length.

Explore More: Proof Hints
9) But how do we know that our conjectures are really true? As you may have seen in the earlier Concurrency Conjecture activity, we must be careful not to form conclusions too easily. Let us investigate the problem further to hopefully come up with some ideas for proofs.
10) Click on 'Show triangle DBC' button, then click on the 'Rotate triangle DBC by -60' button to rotate the interior of △DBC around point B by -60°.
11) What do you notice about the rotated triangle? Try to find other similarly corresponding pairs of triangles.
12) Construct a point at the point of concurrency using the Point Tool on the left. (Scroll down if necessary).
13) Next, measure the six angles formed around the point of concurrency using the Angle Measure Tool on the left. (Scroll down if necessary).
14) What do you notice about the six angles around the point of concurrency? Drag a vertex of △ABC to carefully check your observations.
15) Click on 'Show circumcircle ADB' button to construct the circumcircle of △ADB.
16) What relationship is there between ∠AOB and ∠ADB? What can you conclude from that? (Hint: Look at quadrilateral AOBD.)
17) Press the button that shows all the circumcircles of the equilateral triangles. Look at the other two triangles. What do you notice about these three circles? Check by dragging.

Challenge
18) Now try to use your observations from Questions 10–17 to construct a proof that AE = BF = DC, as well as that lines AE, BF, and DC are concurrent. Discuss your thoughts with a partner. If you get stuck, read the hints in the guided proofs in the accompanying Worksheet.

Further Reflection: Looking Back
The famous mathematician George Polya (1945) emphasized and illustrated the importance of looking back over one's proofs. Now carefully look back and read through your proofs of the two conjectures in 10) for a right triangle.
19) Did your answers to any of the Guiding Questions in the Worksheet use the fact that ∠ABC measures 90°? What does that imply about the two conjectures you just proved?

"A good proof should convey an insight into exactly why the proposition is true.
Such insight sometimes reveals the pleasant, unanticipated surprise that the proposition
is merely a special case of a more general one, thus allowing for its immediate generalization."
— De Villiers, 1998

Generalization
20) In Question 19, you should have noticed that nowhere in the two proofs did we use the given fact that ∠ABC = 90°. What does that imply? Think about it a moment.
21) Since neither of the two proofs depend on ∠ABC = 90°, this means that both proofs would automatically hold for any triangle; in other words, we can deduce that both conjectures hold for any triangle!
22) Click on the 'Link to arbitrary triangle' button to check our conclusion in Q21 above to navigate to a new sketch.
23) In the new sketch use the Line Tool to construct line DC and the Distance Tool on the left to measure the distances DC, EA, and FB. (Alternatively to measuring yourself, click on the 'Show measurements' button to show the line constructions & the distances.) What do you notice? Check your observations by dragging.
24) In the new sketch, measure the six angles formed around the point of concurrency F_T using the Angle Measure Tool on the left - scroll down if necessary. (Alternatively to measuring yourself, click on the 'Show measurements' button to show angle measures.) What do you notice? Check your observations by dragging.
25) What happens if the equilateral triangles are constructed to the interior of △ABC? Do the two results still hold?
(You can check this by dragging say point B over side AC. Can you adapt your proof if they still hold?)
26) Formulate your observations in the form of theorems and write up your proofs.

Discovery Function of Proof
Our discovery that the proof of the concurrency of the lines DC, EA, and FB and DC = EA = FB was independent of ∠ABC = 90°, led to an immediate generalization to any triangle. This is a good example of what has been called the 'discovery' function of proof - whereby proving a result gives insight that leads to its immediate generalization. Earlier examples were already given in the Viviani's theorem as well as in the 'Proof as Discovery' section of Rethinking Proof Activities. Some nice examples at Grade 8 Level are also given in Miyazaki (2000).

Further Generalization
27) Note that the three equilateral triangles constructed on the sides are similar to each other. Can you find arrangements of (more generally) similar triangles on the sides of any triangle ABC such that the concurrency result still holds?
28) The Fermat-Torricelli concurrency result can be generalized as follows:
.....a) click on the 'Link to similar triangles' button to navigate to a new dynamic sketch
.....(to prove this result simply generalize the guided proof given here for equilateral triangles)
.....b) click on the 'Link to similar isosceles △'s' button to navigate to a new dynamic sketch - what result does it reduce to when all the base angles become zero?
.....(to prove this result one can use proof by contradiction, but it's easiest to view it as a special case of Jacobi's theorem)
.....c) click on the 'Link to Jacobi's theorem' button to navigate to a new dynamic sketch
.....(various proofs of this result is possible, but one using Ceva's theorem is given in De Villiers (1995))
.....d) click on the 'Link to further Fermat' button to navigate to a new dynamic sketch which generalizes c) further by subdividing the sides of the base triangle and adding similar triangles to the middle sections
.....(a proof of this result using homothetic triangles is given in De Villiers (1999)).
.....e) Jacobi's theorem can be even further generalized to Anghel's Hexagon Concurrency theorem (click on the 'Link to Anghel's ...' button in the new sketch)
.....(two articles with proofs of this result are given at the above link)

Other Variations
29) What about other regular polygons constructed on the sides of a triangle? Such as regular pentagons, regular septagons, etc.? Are the lines from the apex vertices to the opposite vertices of the base triangle still concurrent? Investigate!
30) And what about regular quadrilaterals with an even number of sides such as regular quadrilaterals (squares), regular hexagons, etc. Can we define a corresponding 'apex' point for regular polygons with an even number of sides constructed on the sides of a triangle so that the lines from these 'apex' points to the directly opposite vertices (of the base triangle) are still concurrent? Investigate!
31) Formulate and explore other variations on your own.
Check: Check your answers to 29) and 30) here: Fermat Torricelli Similar Polygons Concurrency. Can you explain (prove) your observations in terms of the above Similar Isosceles △'s result? (Or Jacobi's theorem)?

Concurrency Special Cases
As you should've seen in 28b) above with similar isosceles △'s above, if we make the base angles zero, we simply obtain the concurrency of the medians of a triangle - see for example the Rethinking Proof activity: Concurrency of Medians.
32) But can we also obtain similar special configurations of Jacobi's theorem for the concurrency of the angle bisectors and the altitudes of a triangle, respectively?
......Indeed we can! As shown by Olivier (1989), if we let DAF, DBE and ECF become straight lines as shown in the 1st figure below, then we obtain the angle bisectors of △ABC. (More-over, the lines DC, EA and FB then become the altitudes of △DEF). Furthermore, as also shown by Olivier (1989) if we reflect △ABC in its sides to obtain the 2nd figure below, then the lines DC, EA and FB become the altitudes of △ABC.
33) Challenge: Can you drag the dynamic sketch of Jacobi's theorem above into these two special cases? Can you also explain why (prove that) the configurations then have the properties mentioned above?

fermat-torricelli special

Generalization of Jacobi's theorem
Jacobi's concurrency theorem can be further generalized to Anghel's Hexagon Concurrency theorem - compare Anghel (2016, 2018).

Historical Note
The Fermat-Torricelli point is named after Pierre de Fermat (1601 – 1665) and Evangelista Torricelli (1608 – 1647) who were both involved in the discovery and proof of the result. A proof that if equilateral triangles DAB, EBC and FCA are constructed on the sides of a triangle ABC, then DC = EA = FB , appears in Mackay (1897), but there may be earlier proofs of this result. More historical information will be provided in the next Building an Airport activity.

Additional Properties of the Configuration
The configuration of a triangle with equilateral or other triangles constructed on its sides is geometrically very rich. In the following two web activities, namely, Building an Airport and Napoleon's Theorem, we will further explore some more properties of this configuration.

References
Anghel, N. (2016). Concurrency and Collinearity in Hexagons. Journal for Geometry and Graphics, Volume 20, No. 2, 159–171.
Anghel, N. (2018). Concurrency in Hexagons - a Trigonometric Tale. Journal for Geometry and Graphics, Volume 22, No. 1, 21–29.
Davis, Hersh, R. (1983). The Mathematical Experience. (Study Edition). Birkhauser.
De Villiers, M. (1995). A Generalisation of the Fermat-Torricelli Point. The Mathematical Gazette, 9(485) (July), pp. 374-378.
De Villiers, M. (1999). A further generalization of the Fermat-Torricelli point. The Mathematical Gazette, 83(496) (March), pp. 14–16.
De Villiers, M. (1999, 2003, 2012). Rethinking Proof with Geometer's Sketchpad (free to download). Key Curriculum Press.
De Villiers, M. (1998). An alternative approach to proof in dynamic geometry. In Lehrer, R.; Chazan, D.(eds). Designing learning environments for developing understanding of geometry and space, (pp. 369-393). NJ:Lawrence Erlbaum.
Jacobi, K. F. A. (1825). Triangulorum Rectilineorum, p. 26.
Mackay, J.S. (1897). Isogonic Centres of a Triangle. Proc. Edinburgh Math. Soc. 15, pp. 100-118.
Miyazaki, M. (2000). What are essential to apply the "discovery" function of proof in lower secondary school mathematics? In Nakahara, T.; Koyama, M. (2000). Proceedings of the 24th Conference of the International Group for the Psychology of Mathematics Education, Vol 4, pp. 1-8, Hiroshima University.
Olivier, A. (1989). 'n Bewys en ander veralgemening van 'n resultaat In De Villiers, M. (1989). Meetkunde, Verklaring en Insig. Pythagoras, No. 21, pp. 33-38.
Park, J. & Flores, A. (2014). Fermat's point from five perspectives. International Journal of Mathematical Education in Science and Technology, DOI: 10.1080/0020739X.2014.979894.
Polya, G. (1945). How to Solve it. Princeton: Princeton University Press.

Other Rethinking Proof Activities
Other Rethinking Proof Activities

External Links
Pierre de Fermat (Wikipedia)
Evangelista Torricelli (Wikipedia)
Fermat point (Wikipedia)
Fermat Points (Wolfram MathWorld)
The Fermat Point and Generalizations (Cut The Knot)
Jacobi's theorem (geometry) (Wikipedia)
Steiner tree problem (Wikipedia)
Soap Bubbles and Mathematics (YouTube video)
X(13) = 1st Isogonic Center (1st Fermat point, Torricelli point, ETC)
X(14) = 2nd Isogonic Center (2nd Fermat point, ETC)
Aimssec Lesson Activities (African Institute for Mathematical Sciences Schools Enrichment Centre)
UCT Mathematics Competition Training Material
SA Mathematics Olympiad Questions and worked solutions for past South African Mathematics Olympiad papers can be found at this link.
(Note, however, that prospective users will need to register and log in to be able to view past papers and solutions.)

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Michael de Villiers, created with WebSketchpad, updated 11; 14; 18; 21 Feb 2026; 21 April 2026.