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 ° it can be proved that △ BC ′ P is a right triangle (extending PB to D, making BD = BP, then PD = PC ′, and ∠ C ′ Pb = 60 ° then △ C ′ PD is an equilateral triangle, which has the property of C ′ B ⊥ BP

1. P is a point on the side BC of △ ABC, and PC = 2bp. We know ∠ ABC = 45 ° and ∠ APC = 60 ° and find the degree of ∠ ACB
2. In RT △ ABC, ∠ BAC = 90 °, ab = AC, M is the middle point of AC, AE ⊥ BM is in E, extending AE to BC is in D, the verification is: ∠ AMB = ∠ CMD
3. In △ ABC, ab = AC, line de intersects AB.BC In D. F, AC extension line in E, and DF = EF, verification: BD = CE
4. P is any point in equilateral △ ABC with side length 1, let t = PA + Pb + PC
Answer to junior high school level, not function, try to isosceles △ thinking

1:
Because PC = 2bp in △ ABC
It is known that ∠ ABC = 45 ° and ∠ APC = 60 °
therefore
∠BAP=5°,
And because PC = 2bp
So:
According to theory 3, it can be concluded that ∠ ACB = 170 degree
2: Provable
But are you too lazy? Do it yourself
3: Provable
But are you too lazy? Do it yourself
4: I don't understand!