Chứng minh BF+CE=BC biết tam giác ABC có góc A=60 độ, tia phân giác của góc B cắt AC tại E

Giải:

Gọi K là giao điểm giữa CF và BE

Kẻ tia phân giác KI của \(\widehat{BKC}\)

\(\Rightarrow\widehat{BKI}=\widehat{CKI}\)

Trong \(\Delta ABC\) có: \(\widehat{A}+\widehat{B}+\widehat{C}=180^o\)

\(\Rightarrow60^o+\widehat{B}+\widehat{C}=180^o\)

\(\Rightarrow\widehat{B}+\widehat{C}=120^o\)

\(\Rightarrow\frac{1}{2}\left(\widehat{B}+\widehat{C}\right)=\frac{1}{2}.120^o\)

\(\Rightarrow\frac{1}{2}\widehat{B}+\frac{1}{2}\widehat{C}=60^o\)

\(\Rightarrow\widehat{B_2}+\widehat{C_1}=60^o\)

Xét \(\Delta BKC\) có: \(\widehat{BKC}+\widehat{B_2}+\widehat{C_1}=180^o\)

\(\Rightarrow\widehat{BKC}+60^o=180^o\)

\(\Rightarrow\widehat{BKC}=120^o\)

Ta có: \(\widehat{B_2}+\widehat{C_1}=\widehat{FKB}\)

\(\Rightarrow\widehat{FKB}=60^o\)

Mà \(\widehat{FKB}=\widehat{EKC}\) ( đối đỉnh )

\(\Rightarrow\widehat{EKC}=60^o\)

Xét \(\Delta FKB,\Delta IKB\) có:
\(\widehat{B_1}=\widehat{B_2}\left(=\frac{1}{2}\widehat{B}\right)\)

BK: cạnh chung

\(\widehat{FKB}=\widehat{IKB}\left(=60^o\right)\)

\(\Rightarrow\Delta FKB=\Delta IKB\left(g-c-g\right)\)

\(\Rightarrow BF=BI\) ( cạnh t/ứng )

Xét \(\Delta EKC,\Delta IKC\) có:

\(\widehat{C_1}=\widehat{C_2}\left(=\frac{1}{2}\widehat{C}\right)\)

KC: cạnh chung

\(\widehat{EKC}=\widehat{IKC}\left(=60^o\right)\)

\(\Rightarrow EC=IC\) ( cạnh t/ứng )

Có: \(BI+IC=BC\)

\(\Rightarrow BF+CE=BC\)

\(\Rightarrowđpcm\)