Problem
GEO-B3-M05-P015 Triangle of Second Intersections
Let \(P\) be the first Brocard point of \(ABC\): \(\angle ABP=\angle BCP=\angle CAP\). Lines \(AP,BP,CP\) meet the circumcircle again at \(A_1,B_1,C_1\). Prove that triangle \(A_1B_1C_1\) is congruent to triangle \(BCA\).
C. Hint 1. Convert the equal Brocard angles into equal arcs of the circumcircle.
D. Hint 2. Compare the arcs subtending \(A_1B_1,B_1C_1,C_1A_1\).
Let the common Brocard angle be \(\varphi\). Since \(A,P,A_1\), \(B,P,B_1\), \(C,P,C_1\) are collinear, we get \(\angle CAA_1=\angle ABB_1=\angle BCC_1=\varphi\). Hence the corresponding directed arcs satisfy \(\widehat{CA_1}=\widehat{AB_1}=\widehat{BC_1}=2\varphi\).
Now compare the arcs between the new points. Passing from arc \(BC\) to arc \(A_1B_1\), both endpoints are shifted along the circle by the same directed arc \(2\varphi\). Therefore \(\widehat{A_1B_1}=\widehat{BC}\). Similarly, \(\widehat{B_1C_1}=\widehat{CA}\) and \(\widehat{C_1A_1}=\widehat{AB}\). Equal arcs give equal chords, so \(\triangle A_1B_1C_1\cong \triangle BCA\).
Watch notation carefully: this is an oriented-arc problem.