... : BC : CA = A 1 B 1 : B 1 C 1 : C 1 A 1 ;b) AB : BC = A 1 B 1 : B 1 C 1 and ∠ABC = ∠A 1 B 1 C 1 ;c) ∠ABC = ∠A 1 B 1 C 1 and ∠BAC = ∠B 1 A 1 C 1 .2) Triangles AB 1 C 1 and AB2C2cut off from ... A 1 B 1 C 1 , i.e., thecenter of the circumscribed circle of triangle ABC. Since A 1 H ⊥ B 1 C 1 and B 1 H ⊥ A 1 C 1 ,it follows that ∠(A 1 H, HB 1 ) = ∠(B 1 C 1 , A 1 C 1 ) = ∠(A 1 C, CB 1 ), ... ANGLESHence,∠(A 1 C 1 , C 1 B 1 ) = ∠(AC, CB) = ∠(AC, CD) + ∠(DC, CB)= ∠(AB 1 , B 1 D) + ∠(DA 1 , A 1 B) = ∠(A 1 D, DB 1 ),i.e., points A 1 , B 1 , C 1 and D lie on one circle. Therefore, ∠(A 1 C 1 , C 1 B)...