7. Friendship of Kiepert perspectors
Given any real number t, Let Xt and Yt be the points that divide CBc and BCb such that CXt : CBc = BYt : BCb = t :1, and let Mt be their midpoint. Then BCMt is an isosceles triangle, with base angle arctant = ∠BAY t. See Figure 5. Extend AXt to X t on BaBc, and AYt to Y t on CaCb and let M t be the midpoint of X tY t . Then BaCaM t is an isosceles triangle, with base angle arctan1 t = ∠Y t ACa = π 2 −∠BAYt. Also, by the similarity of triangles AXtYt and AX tY t