As shown in the figure, at the isosceles right angle △ ABC, ∠ C = 90 °, AC = BC, D is any point on AB, AE ⊥ CD is at e, BF ⊥ CD intersects the extension line of CD at F, CH ⊥ AB is at h, Give AE to g, verify: CE = BF

As shown in the figure, at the isosceles right angle △ ABC, ∠ C = 90 °, AC = BC, D is any point on AB, AE ⊥ CD is at e, BF ⊥ CD intersects the extension line of CD at F, CH ⊥ AB is at h, Give AE to g, verify: CE = BF

It is proved that: ∵ AE ⊥ CD, BF ⊥ CD ∥ AE ∥ BF ∥ FBA = - EAB ∥ AC = BC, ∥ C = 90 ∥ cab = - CBA = 45 ∥ ch ⊥ HB ∥ EAB + ∥ AGH = 90 ∥ AE ⊥ CD ∥ HCE + ∥ CGE = 90 ∥ CGE = - AGH ∥ EAB = - HCE = - FBA ∥ ch ⊥ Hb, AC = BC, ∥ C = 90 ∥ BCH

Explore the following relationship between ∠ A and ∠ P and explain the reasons
(1) As shown in Figure 1, BP and CP divide ∠ ABC and ∠ ACB equally; (2) as shown in Figure 2, BP and CP divide the complementary angle of ∠ ABC and ∠ ACB equally; (3) as shown in Figure 3, BP divides the complementary angle of ∠ ABC equally and CP divides the complementary angle of ∠ ACB equally

(1) The BP and CP, BP and CP, respectively, bisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbisbis \ \\\ \ \\\\\\\\\\\\\\\\\c + ∠ PCA +As a result, there are three cases of the following 12; (3) (3)); (3); (3); (3); (3); (3) the DBC (DBC) is a + ACB, and the ECB is a + ABC, ABC, ABC, ABC, ABC, BP, and CP are the equal division of DBC and ECD, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, and respectively, respectively, respectively, respectively, respectively, and respectively, respectively, respectively, respectively, respectively, respectively, respectively, and respectively, respectively, respectively, respectively, respectively, and respectively, respectively, respectively, respectively, respectively, respectively, respectively, and respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, the DBC and the DBC and ECD, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively, respectively(a) P = 90 ° + 12 ∠ a ；（2）∠P=∠A；（3）∠P=90°-12∠A，

As shown in the figure, in the above question, if BP and CP are bisectors of ∠ CBD and ∠ BCE respectively, what is the relationship between ∠ P and ∠ a? Try to prove your conclusion
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Because BP and CP are bisectors, so ∠ DBP = ∠ PBC, ∠ BCP = ∠ PCE, because (180-2 ∠ PBC) + (180-2 ∠ BCP) = 180 - ∠ a, that is 360-2 (∠ PBC + ∠ BCP) = 180 - ∠ A and ∠ PBC + ∠ BCP = 180 - ∠ p, we get 2 ∠ P + ∠ a = 180 ∠

Point a is inside the circle, point B is outside the circle, and C and D are on the circle

Let BC and circle intersect at point E
∵∠ CED is the outer angle of the triangle bed, which is larger than the corresponding inner angle ∠ CBD
That is ∠ CED > CBD
A is in the circle, e is on the circle, and arc CD is the common arc of ∠ CAD and ∠ CED
∴∠CAD>∠CED ②
From (1) and (2), we can get ∠ CAD > ∠ CBD