\(S_{\Delta ABD}=\dfrac{80.60}{2}=2400\)

Applying the Pythagoras theorem to the right \(\Delta ABD\),we have :

\(BD=\sqrt{AB^2+AD^2}=\sqrt{80^2+60^2}=\sqrt{10000}=100\)

\(\Rightarrow AH=\dfrac{2S_{\Delta ABD}}{BD}=\dfrac{4800}{100}=48\)

Sorry for the wrong answer, but Phan Thanh Tinh made the right comment.