Proof of Lemma 3.
For an equilateral triangle ABC the envelope of its Wallace-Simson
lines is a tricuspidal hypocycloid whose vertices are the vertices of an
equilateral triangle concentric with ABC, with sides parallel to those
of ABC and whose size is 3/2 that of ABC.

The proof is easily obtained from the two following remarks:

Figure 1
(a) If a rectangle with diagonal OP (diagonal length  2m) is enlarged with two rectangles as the figure shows, then one has the indicated relations between the angles.

Figure 2

    (b) To the elements of Figure 1 one has added in Figure 2 the circles U and V, with center at O and radii 2m and 3m, the equilateral triangle ABC, the circle W with center at P and radius m, that intersects QJ at T and L.
    Then, since the angle PLT is 3t/2 (according to Figure 1), we have that the angle SPT is 3t and thus T is a point of the hypocycloid generated by W when it rolls inside V starting from the position in which the center of W is on the line OA.
    Since the angle STQ is 90º and ST is instantaneous rotation radius of the circle W when it rolls inside V, the line TQ is tangent at T to the hypocycloid. On the other hand since angle PQT is  t/2 we have (see Figure 3 below) that QT  is the Wallace-Simson line of P with respect to  ABC. Thus the Wallace-Simson line of P with respect to ABC is tangent to the hypocycloid at the point T. This proves the lemma.
Figure 3
    Since BPQR is a cyclic quadrilateral, the angles PBR and PQR are equal. On the other hand the angle PBR, inscribed in the circle U, is half the angle POA. Thus the Wallace-Simson line of P, i.e. QR, is at an angle t/2 with PQ and so R coincides with the point T of the previous figure 2.