Ta có: \(\overrightarrow {{\rm{AJ}}} = \frac{1}{2}\left( {\overrightarrow {AH} + \overrightarrow {AC} } \right);\overrightarrow {BI} = \frac{1}{2}\left( {\overrightarrow {BA} + \overrightarrow {BH} } \right)\)

\(\overrightarrow {AJ} .\overrightarrow {BI} = \frac{1}{4}\left( {\overrightarrow {AH} + \overrightarrow {AC} } \right)\left( {\overrightarrow {BA} + \overrightarrow {BH} } \right) = \frac{1}{4}\left( {\overrightarrow {AH} .\overrightarrow {BA} + \overrightarrow {AH} .\overrightarrow {BH} + \overrightarrow {AC} .\overrightarrow {BA} + \overrightarrow {AC} .\overrightarrow {BH} } \right)\)

          \( = \frac{1}{4}\left( {\overrightarrow {AH} .\overrightarrow {HA} + 0 + 0 + \overrightarrow {HC} .\overrightarrow {BH} } \right) = \frac{1}{4}\left( { - A{H^2} + BH.HC} \right)\)

Vì ∆ABC vuông tại A nên: \(A{H^2} = HB.HC\)

Do đó: \(\overrightarrow {AJ} .\overrightarrow {BI} = 0 \Leftrightarrow AJ \bot BI\).