... ∠BPC·sin ∠QACsin ∠PAQ·sin ∠QCPsin ∠QCA= 1, which is equivalent tosin ∠AP Bsin ∠BPC=sin ∠QCAsin ∠QAC2because ∠QCA = ∠PAQ and ∠QAC = ∠QCP . Denote by S(XY Z)thearea of a triangle ... in the previous case we obtain (2 +√3)2n+1= −1; i.e., the orderof 2 +√3 in the multiplicative group Zp(√3)∗is 2n+2. We cannotfinish the proof as in the previous case: in fact, ... segments BC, CA, AB.Denote by a, b, c the length of the segments BC, CA, AB. It follows thata = z + y, b = z + x, c = x + y,soD, E, F are the points where theexcircles touch the sides of ABC....