Triangles (8) ... Proof by Ceva's Theorem of concurrence at the Fermat Point

Ceva's Theorem (and its converse) were covered in a previous post - Ceva's Theorem. 

In the earlier post on the Fermat Point, it was stated there is a concurrency at the Fermat Point but no proof was presented. As is often the case, there are various proofs. This post offers a proof using the converse of Ceva's Theorem.

Steps 1 and 2 establish two mathematical results which will be used in the proof.

Step 1

△ ABC is equilateral. AP is a cevian.  Let ∠ BAP = a and ∠ CAP = b. Then

Because △ ABC is equilateral, AB = AC. Also,  angles APB and APC are complementary and so their sines are equal.

Step 2

△ ABC is any acute triangle. AH is the altitude from A to side BC. Then 

 

Step 3

The next diagram shows a triangle ABC with equilateral triangles drawn on each side - △ APB, △  BQC, △ ARC

Straightlines join P to C, Q to A and R to B. The angles made by PC at point P are P1 and P2. AQ makes angles Q1 and Q2 at point Q. BR makes angles R1 and R2 at point R.

Applying Step and from △ BQC. △ ARC, △ APB we obtain, respectively

 =

 =

 =

and so 

 x  x  =  x  x

Step 4

Various pairs of triangles are congruent and that makes it possible to find all further angles which are shown in the next diagram. Congruent triangle pairs are △ ABQ ≅ △ PCB and △ AQC ≅ △ RBC and △ ABR ≅ △ PAC.

By applying Step 2 (above) we find that

From △ ABQ

 =  =      because △ BCQ is equilateral, BQ = BC

From △ ACQ

 =  =      because △ BCQ is equilateral, CQ = BC

From △ ABR

 =  =      because △ ARC is equilateral, AR = AC

Step 5

Take the expression arrived at in Step 3

 x  x  =  x  x

This can be rearranged to

 x  x  and, applying Step 4, that is equal to  x  x

And that is equal to 1. Thus, by the converse of Ceva’s theorem, there is concurrence at a single point – the Fermat Point.

Not the most straightforward proof but a useful demonstration of how Ceva's Theorem can be applied.