từ A kẻ đường cao AD.
Ta có SABC =SACD +SACD . Do đó
\(\frac{1}{2}\).AB.AC.\(\sin\)60
=\(\frac{1}{2}\)AB.AD.sin30 +\(\frac{1}{2}\)AC.AD.sin30
Suy ra AB.AC.\(\sqrt{3}\)=AB.AD.\(\frac{1}{2}\) + AC.AD.\(\frac{1}{2}\)
Hay AB.AC.\(\sqrt{3}\)= AD(AB+AC)
nên \(\frac{\sqrt{3}}{AD}\)=\(\frac{AB+AC}{AB.AC}\)=\(\frac{1}{AC}\) + \(\frac{1}{AB}\) (đpcm).
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