... =∠(XB2, B2M) + ∠(A1N, XA1) + 90 ◦.Since∠(XB2, B2M) + ∠(AC, CB) + ∠(NA1, A1X) + 90 ◦= 0◦,we have: ∠(MN, XN) + ∠(AC, CB) = 0◦.SOLUTIONS 1 29 5 .94 . If point R on the given circle ... themidpoints of the corresponding heights intersect at Lemoine’s point.See also Problems 11.22, 19. 54, 19. 55.112 CHAPTER 5. TRIANGLESProblems for independent study5.136. Prove that the projection ... thistriangle.5.3. Clearly,∠BOC = 180◦− ∠CBO − ∠BCO = 180◦−∠B2−∠C2= 90 ◦+∠A2and ∠BOaC = 180◦− ∠BOC, because ∠OBOa= ∠OCOa= 90 ◦.5.4. Let AA1, BB1and CC1be the bisectors of triangle...