You can move the points A, B and C (click on the point and drag it).
Press the keys “+” and “−” to zoom in or zoom out the visualization window and use the arrow keys to translate it.
You can also construct all centers related with this one (as described in ETC) using the “Run Macro Tool”. To do this, click on the icon , select the center name from the list and, then, click on the vertices A, B and C successively.
|Information from Kimberling's Encyclopedia of Triangle Centers|
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)