a) Ta có tam giác vuông ABC (do AD vuông góc BC) nên ta có:
\[AB^2 + AC^2 = BC^2\]
\[AB^2 + (AC - AB)^2 = BC^2\]
\[AB^2 + AC^2 - 2AB \cdot AC + AB^2 = BC^2\]
\[2AB^2 - 2AB \cdot AC = BC^2 - AC^2\]
\[2AB(AB - AC) = BC^2 - AC^2\]
\[2AB \cdot BC = BC^2 - AC^2\]
\[AB = \frac{BC^2 - AC^2}{2BC}\]
\[AB = \frac{9^2 - 4^2}{2 \cdot 9}\]
\[AB = \frac{81 - 16}{18}\]
\[AB = \frac{65}{18}\]

Xét tam giác vuông AID, ta có:
\[\frac{AI}{AB} = \frac{ID}{IB}\]
\[\frac{AI}{\frac{65}{18}} = \frac{AD}{4}\]
\[AI = \frac{65}{18} \cdot \frac{4}{AD}\]
\[AI = \frac{260}{18 \cdot AD}\]
\[AI = \frac{130}{9 \cdot AD}\]

Với tam giác vuông ABC, ta có:
\[\frac{AC}{AB} = \frac{BC}{AC}\]
\[\frac{AC}{\frac{65}{18}} = \frac{9}{AC}\]
\[AC^2 = \frac{65 \cdot 9}{18}\]
\[AC = \sqrt{\frac{585}{18}}\]
\[AC = \frac{\sqrt{585}}{3}\]

Xét tam giác vuông AIC, ta có:
\[\frac{AI}{AC} = \frac{IC}{AC}\]
\[\frac{AI}{\frac{\sqrt{585}}{3}} = \frac{9}{\frac{\sqrt{585}}{3}}\]
\[AI = \frac{9 \cdot \sqrt{585}}{3}\]
\[AI = 3 \cdot \sqrt{585}\]
\[AI = 3 \cdot \sqrt{3 \cdot 195}\]
\[AI = 3 \cdot 3 \cdot \sqrt{195}\]
\[AI = 9 \cdot \sqrt{195}\]

Vậy ta có:
\[AD = \frac{130}{9 \cdot AI}\]
\[AD = \frac{130}{9 \cdot 9 \cdot \sqrt{195}}\]
\[AD = \frac{130}{81 \cdot \sqrt{195}}\]
\[AD = \frac{130 \cdot \sqrt{195}}{81 \cdot 195}\]
\[AD = \frac{130 \cdot \sqrt{195}}{15795}\]

b) Ta có:
\[\angle BEF = \angle BEH + \angle HEF\]
\[\angle BEF = 90^\circ + 90^\circ\]
\[\angle BEF = 180^\circ\]

Vậy tam giác BEF là tam giác cân.