Đặt \(AB' = mAB,AC' = nAC\). Ta có:

\(\begin{array}{l}\overrightarrow {AM}  = \dfrac{1}{2}\left( {\overrightarrow {AB}  + \overrightarrow {AC} } \right) = \dfrac{1}{2}\overrightarrow {AB}  + \dfrac{1}{2}\overrightarrow {AC} \\\overrightarrow {B'C'}  = \overrightarrow {AC'}  - \overrightarrow {AB'}  = n\overrightarrow {AC}  - m\overrightarrow {AB} \\ \Rightarrow \overrightarrow {AM} .\overrightarrow {B'C'}  = \left( {\dfrac{1}{2}\overrightarrow {AB}  + \dfrac{1}{2}\overrightarrow {AC} } \right)\left( {n\overrightarrow {AC}  - m\overrightarrow {AB} } \right)\\ = \dfrac{n}{2}\overrightarrow {AB} .\overrightarrow {AC}  + \dfrac{n}{2}A{C^2} - \dfrac{m}{2}A{B^2} - \dfrac{m}{2}\overrightarrow {AC} .\overrightarrow {AB} \\ = \dfrac{n}{2}.0 + \dfrac{1}{2}\left( {nA{C^2} - mA{B^2}} \right) - \dfrac{m}{2}.0\\ = \dfrac{1}{2}\left( {nA{C^2} - mA{B^2}} \right)\end{array}\)

Mà \(AB.AB' = AC.AC' \Rightarrow AB.mAB = AC.nAC \Leftrightarrow mA{B^2} - nA{C^2} = 0\)

Do đó \(\dfrac{1}{2}\left( {nA{C^2} - mA{B^2}} \right) = 0\) hay \(AM \bot B'C'\).

Mong bạn thông cảm nếu không thấy rõ