MathDB

2

Part of 2009 China Northern MO

Problems(1)

angle bisector and BE+CF=EF wanted, cos B+ cos C=1, <ABF =< ACE = 90^o

Source: China Northern MO 2009 p2 CNMO

12/12/2020
In an acute triangle ABCABC , AB>ACAB>AC , cos⁡B+cos⁡C=1 \cos B+ \cos C=1 , E,FE,F are on the extend line of AB,ACAB,AC such that ∠ABF=∠ACE=90\angle ABF = \angle ACE = 90 . (1) Prove :BE+CF=EFBE+CF=EF ; (2) Assume the bisector of ∠EBC\angle EBC meet EFEF at PP , prove that CPCP is the bisector of ∠BCF\angle BCF. https://cdn.artofproblemsolving.com/attachments/a/2/c554c2bc0b4e044c45f88138568f5234d544a8.png
geometryangle bisectorright angletrigonometry