As shown in the figure, in △ ABC, ab = AC, ∠ BAC = 80 °, point P is within △ ABC, ∠ PBC = 10 °, PCB = 30 °, calculate the degree of ∠ BAP What are you doing?

As shown in the figure, in △ ABC, ab = AC, ∠ BAC = 80 °, point P is within △ ABC, ∠ PBC = 10 °, PCB = 30 °, calculate the degree of ∠ BAP What are you doing?

(1) Right triangle area = AB / 2 = ch / 2, that is ab = ch Pythagorean theorem: C ^ 2 = a ^ 2 + B ^ 2 (c + H) ^ 2 = C ^ 2 + 2CH + H ^ 2 = a ^ 2 + B ^ 2 + 2Ab + H ^ 2 = (a + b) ^ 2 + H ^ 2

In △ ABC, ab = AC, ∠ BAC = 80 °, P in △ ABC, ∠ PBC = 10 °, PCB = 20 °, then the degree of ∠ PAB is ()
A. 50°B. 60°C. 70°D. 65°

As shown in the figure, make the symmetry point P ′ of P about AC, connect AP ′, P ′ C, PP ′, then p ′ C = PC, ACP ′ = ∠ ACP. ∵ AB = AC, ∵ BAC = 80 °, and ∵ ABC = ∵ ACB = 50 °, and