In △ ABC, ad is an angular bisector, BP ⊥ ad is at point P, and ab = 5, BP = 2, AC = 9. It is proved that: ∠ ABC = 3 ∠ C

In △ ABC, ad is an angular bisector, BP ⊥ ad is at point P, and ab = 5, BP = 2, AC = 9. It is proved that: ∠ ABC = 3 ∠ C

Extend BP to AC to e
Because ad is equal to BAC, BP is vertical to AD
Then: ab = AE = 5, EC = ac-ae = 4
Angle ABP = angle AEB
BE=2BP=4
So: be = EC
The triangle BEC is isosceles triangle
Angle c = angle EBC
Angle AEB = angle c + angle EBC = 2 angle c
So: angle ABC = angle ABP + angle EBC = angle AEB + angle c = 3 angle c

As shown in the figure, in △ ABC, ad is the bisector of ∠ BAC, BP ⊥ ad, and the perpendicular foot is p. it is known that ab = 5, BP = 2, AC = 9

It is proved that: extending BP, crossing AC to e, ∵ ad, bisecting ∵ BAC, BP ⊥ ad, ∵ BAP = ∵ EAP, ∵ APB = ∵ ape, and ∵ AP = AP, ≌ ABP ≌ AEP, ≌ BP = PE, AE = AB, ∵ AEB = ∵ Abe, ∵ be = BP + PE = 4, AE = AB = 5, ? CE = ac-ae = 9-5 = 4, ≌ CE = be, ≌ BCE are isosceles triangles