Question

Cho biết \(\dfrac{AB'}{AB}=\dfrac{AC'}{AC}\left(h.6\right)\)

Chứng minh rằng : 

a) \(\dfrac{AB'}{B'B}=\dfrac{AC'}{C'C}\)

b) \(\dfrac{BB'}{AB}=\dfrac{CC'}{AC}\)

Hướng dẫn : Áp dụng tính chất của tỉ lệ thức 

Answer 1

a) Ta có:

$\frac{A{B}^{\prime }}{AB}$ = $\frac{A{C}^{\prime }}{AC}$ => $\frac{AC}{A{C}^{\prime }}$ = $\frac{AB}{A{B}^{\prime }}$

=> $\frac{AC}{A{C}^{\prime }}$ - 1 = $\frac{AC-A{C}^{\prime }}{A{C}^{\prime }}$ = $\frac{AB-A{B}^{\prime }}{A{B}^{\prime }}$

=> $\frac{C{C}^{\prime }}{A{C}^{\prime }}$ = $\frac{B^{\prime }B}{A{B}^{\prime }}$ => $\frac{A{B}^{\prime }}{B{B}^{\prime }}$ = $\frac{A{C}^{\prime }}{C{C}^{\prime }}$

b) Vì $\frac{A{B}^{\prime }}{AB}$ = $\frac{A{C}^{\prime }}{AC}$ mà AB' = AB - B'B, AC' = AC - C'C.

Answer 2

a) Ta có:

\(\dfrac{AB'}{AB}=\dfrac{AC'}{AC}\Rightarrow\dfrac{AB}{AC}=\dfrac{AB'}{AC'}\)

Áp dụng tc dãy tỉ số bằng nhau ta có;

\(\dfrac{AB}{AC}=\dfrac{AB'}{AC'}=\dfrac{AB-AB'}{AC-AC'}=\dfrac{BB'}{CC'}\)

\(\Rightarrow\dfrac{AB'}{AC'}=\dfrac{BB'}{CC'}\Leftrightarrow\dfrac{AB'}{BB'}=\dfrac{AC'}{CC'}\)

b) Ta có:

\(\dfrac{AB}{AC}=\dfrac{BB'}{CC'}\left(cmt\right)\)

\(\Leftrightarrow\dfrac{AB}{BB'}=\dfrac{AC}{CC'}\Leftrightarrow\dfrac{BB'}{AB}=\dfrac{CC'}{AC}\)