Fermat Point - Concurrency

Concurrency

Here is a proof using properties of concyclic points to show that the three red lines in Fig 1 are concurrent and cut one another at angles of 60°.

In Fig 3 suppose RC and BQ intersect at F, and two lines, AF and FP, are drawn. We aim to prove that the points A,F,P are collinear.

The triangles RAC and BAQ are congruent because the second is a 60° rotation of the first about A. Hence ∠ARF = ∠ABF and ∠AQF = ∠ACF. By converse of angle in the same segment, ARBF and AFCQ are both concyclic. Thus ∠AFB = ∠AFC = ∠BFC = 120°. Because ∠BFC and ∠BPC add up to 180°, BPCF is also concyclic. Hence ∠BFP = ∠BCP = 60°. Because ∠BFP + ∠BFA = 180°, AFP is a straight line.

Q.E.D.

This proof only applies in Case 2 since if ∠BAC > 120° A lies inside the circumcircle of BPC which switches the relative positions of A and F. However it is easily modified to cover Case 1. Then ∠AFB = ∠AFC = 60° hence ∠BFC = ∠AFB = ∠AFC = 120° which means BPCF is concyclic so ∠BFP = ∠BCP = 60° = ∠BFA. Therefore A lies on FP.

Clearly any 3 lines perpendicular to the red ones in Fig 1, in particular those joining the centres of the circles, must also cut at angles of 60° and thereby form an equilateral triangle. This curiosity is known as Napoleon's Theorem.