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 \(AM \bot B'C'\).
