As shown in the figure, it is known that P is a point on the side BC of △ ABC, and PC = 2PB. If ∠ ABC = 45 ° and ∠ APC = 60 °, find the size of ∠ ACB

As shown in the figure, it is known that P is a point on the side BC of △ ABC, and PC = 2PB. If ∠ ABC = 45 ° and ∠ APC = 60 °, find the size of ∠ ACB

If we make the symmetry point C 'of C with respect to AP and connect AC', BC 'and PC', then PC ′ = PC = 2PB, ∠ APC ′ = ∠ APC = 60 ° we can prove that △ BC ′ P is a right triangle (extend PB to D, make BD = BP, then PD = PC ′, and ∠ C ′ Pb = 60 °, then △ C ′ PD is an equilateral triangle. By the property of three lines in one, there is C ′ B ⊥ BP, ∠ C ′ BP = 90 ° because ∠ ABC = 45 °, so ∠ C ′ Ba = 45 ° = ∠ ABC, so Ba is flat The distance from a to BC 'is equal to the distance from a to BC, and the distance from a to PC' is equal to the distance from a to PC '(i.e. BC), so the distance from a to BC' is equal to the distance from a to PC ', so a is the point on the angular bisector, that is, the point on the angular bisector where C' a bisects ∠ MC 'p with ∠ AC' p = 12 ∠ MC 'p = 75 °= ∠ ACB