Trilinears           sec A/2 : sec B/2 : sec C/2
                                    = f(a,b,c) : f(b,c,a) : f(c,a,b), where f(a,b,c) = [bc/(b + c - a)]^1/2

Barycentrics    sin A/2 : sin B/2 : sin C/2

In notes dated 1987, Yff raises a question concerning certain triangles lying within ABC: can three isoscelizers (as defined in connection with X(173), P(B)Q(C), P(C)Q(A), P(A)Q(B) be constructed so that the four triangles P(A)Q(A)A, P(B)Q(B)B, P(C)Q(C)C, ABC are congruent? After proving that the answer is yes, Yff moves the three isoscelizers in such a way that the three outer triangles, P(A)Q(A)A, P(B)Q(B)B, P(C)Q(C)C stay congruent and the inner triangle, ABC, shrinks to X(174).

X(174) lies on these lines: 1,167    2,236    7,234    57,173    175,483    176,1143    188,266    481,1127    558,1489

X(174) = isogonal conjugate of X(259)
X(174) = anticomplement of X(2090)
X(174) = X(508)-Ceva conjugate of X(188)
X(174) = cevapoint of X(I) and X(J) for these (I,J): (1,173), (259,266)
X(174) = X(I)-cross conjugate of X(J) for these (I,J): (1,1488), (177,7), (259,188)
X(174) = crosssum of X(1) and X(503)
X(174) = X(556)-beth conjugate of X(556)