設正方形 $ABCD$ 的邊長為 $s$，$\overline{BE} = h$ 垂直於底面。以 $B$ 為原點建立空間坐標系：
$$B = (0,0,0),\ A = (s,0,0),\ C = (0,s,0),\ D = (s,s,0),\ E = (0,0,h).$$
2. 在 $\triangle AEB$ 中，因為 $\angle ABE = 90^\circ$，所以
$$\cot\angle AEB = \frac{\overline{BE}}{\overline{AB}} = \frac{h}{s} = \frac{2\sqrt{6}}{5}.$$
3. 在 $\triangle CED$ 中（以 $E$ 為頂點）：$\overset{\large\rightharpoonup}{EC} = (0, s, -h)$，$\overset{\large\rightharpoonup}{ED} = (s, s, -h)$。因為
$$\overset{\large\rightharpoonup}{EC} \cdot \overset{\large\rightharpoonup}{ED} = 0 \cdot s + s \cdot s + (-h)(-h) = s^2 + h^2,$$
$$|\overset{\large\rightharpoonup}{EC}| = \sqrt{s^2 + h^2}, \; |\overset{\large\rightharpoonup}{ED}| = \sqrt{2s^2 + h^2},$$
所以
$$\cos\angle CED = \frac{s^2 + h^2}{\sqrt{s^2 + h^2} \cdot \sqrt{2s^2 + h^2}} = \frac{\sqrt{s^2 + h^2}}{\sqrt{2s^2 + h^2}}.$$
4. 計算 $\sin\angle CED$：由向量叉積得
$$|\overset{\large\rightharpoonup}{EC} \times \overset{\large\rightharpoonup}{ED}| = |(0,s,-h) \times (s,s,-h)| = |(sh + sh, -sh, -s^2)| = s\sqrt{h^2 + s^2}.$$
所以
$$\sin\angle CED = \frac{s\sqrt{s^2 + h^2}}{\sqrt{s^2 + h^2} \cdot \sqrt{2s^2 + h^2}} = \frac{s}{\sqrt{2s^2 + h^2}}.$$
5. 因此
$$\cot\angle CED = \frac{\cos\angle CED}{\sin\angle CED} = \frac{\sqrt{s^2 + h^2}/\sqrt{2s^2 + h^2}}{s/\sqrt{2s^2 + h^2}} = \frac{\sqrt{s^2 + h^2}}{s} = \sqrt{1 + \left(\frac{h}{s}\right)^2}.$$
6. 代入 $\frac{h}{s} = \frac{2\sqrt{6}}{5}$：
$$\cot\angle CED = \sqrt{1 + \frac{24}{25}} = \sqrt{\frac{49}{25}} = \frac{7}{5}.$$
故填 $\frac{7}{5}$。
