Question

[NGUYỄN THÚY AN]

Cho\(\Delta\) ABC có AB=4, BC=5, AC=6. Kẻ phân giác BE

1/ tính độ dài EA, EC, EB

2/ CM \(\widehat{B}\)= 2\(\widehat{C}\)

3/ Trên tia đối tia BA lấy điểm F sao cho BF=5. Tính CF

Answer 1

[Pham Van Hung]

a,Tam giác ABC có BE là tia phân giác của \(\widehat{ABC}\left(gt\right)\Rightarrow\frac{BA}{BC}=\frac{EA}{EC}\)

\(\Rightarrow\frac{EA}{EC}=\frac{4}{5}\Rightarrow\frac{EA}{4}=\frac{EC}{5}=\frac{EA+EC}{4+5}=\frac{AC}{9}=\frac{6}{9}=\frac{2}{3}\)

\(\Rightarrow EA=\frac{8}{3}\left(cm\right),EC=\frac{10}{3}\left(cm\right)\)

Ta có: \(\frac{AB}{AE}=\frac{4}{\frac{8}{3}}=\frac{3}{2}\)

\(\frac{AC}{AB}=\frac{6}{4}=\frac{3}{2}\Rightarrow\frac{AB}{AE}=\frac{AC}{AB}\)

\(\Delta ABC\infty\Delta AEB\left(c.g.c\right)\Rightarrow\frac{AC}{AB}=\frac{BC}{EB}\Rightarrow\frac{6}{4}=\frac{5}{EB}\Rightarrow EB=\frac{10}{3}\left(cm\right)\)

b, \(\Delta ABC\infty\Delta AEB\left(cmt\right)\Rightarrow\widehat{ACB}=\widehat{ABE}\)

Mà BE là tia p/g của \(\widehat{ABC}\left(gt\right)\Rightarrow\widehat{ABC}=2\widehat{ABE}\Rightarrow\widehat{ABC}=2\widehat{ACB}\)

c, \(\Delta BCF\) cân tại B (vì BC = BF = 5 cm) \(\Rightarrow\widehat{F}=\widehat{BCF}\)

Do đó: \(\widehat{ABE}=\frac{1}{2}\widehat{ABC}=\frac{1}{2}\left(\widehat{BCF}+\widehat{F}\right)=\widehat{F}\)

\(\Rightarrow BE//FC\Rightarrow\frac{BE}{FC}=\frac{AB}{AF}\Rightarrow\frac{\frac{10}{3}}{FC}=\frac{4}{9}\Rightarrow FC=7,5\left(cm\right)\)