我們將向量分解以進行推導。已知 $\overset{\large\rightharpoonup}{AD} \cdot \overset{\large\rightharpoonup}{AB} = 0$ 且 $\overset{\large\rightharpoonup}{AD} \cdot \overset{\large\rightharpoonup}{AC} = 0$。 將 $\overset{\large\rightharpoonup}{DB}$ 與 $\overset{\large\rightharpoonup}{DC}$ 表示為： - $\overset{\large\rightharpoonup}{DB} = \overset{\large\rightharpoonup}{AB} - \overset{\large\rightharpoonup}{AD}$ - $\overset{\large\rightharpoonup}{DC} = \overset{\large\rightharpoonup}{AC} - \overset{\large\rightharpoonup}{AD}$ 1. 計算內積： $$\overset{\large\rightharpoonup}{DB} \cdot \overset{\large\rightharpoonup}{DC} = (\overset{\large\rightharpoonup}{AB} - \overset{\large\rightharpoonup}{AD}) \cdot (\overset{\large\rightharpoonup}{AC} - \overset{\large\rightharpoonup}{AD}) = \overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AC} - \overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AD} - \overset{\large\rightharpoonup}{AD} \cdot \overset{\large\rightharpoonup}{AC} + |\overset{\large\rightharpoonup}{AD}|^2$$ 由於垂直條件，簡化為： $$\overset{\large\rightharpoonup}{DB} \cdot \overset{\large\rightharpoonup}{DC} = \overline{DA}^2 + \overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AC}$$ 故選項 $(1)$ 為負號 $-\overline{DA}^2$，是錯誤的。 2. 檢驗角度： - 若 $\angle BAC$ 是直角 $\implies \overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AC} = 0 \implies \overset{\large\rightharpoonup}{DB} \cdot \overset{\large\rightharpoonup}{DC} = \overline{DA}^2 > 0$，故 $\angle BDC$ 為銳角。$(2)$ 錯誤。 - 若 $\angle BAC$ 是銳角 $\implies \overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AC} > 0 \implies \overset{\large\rightharpoonup}{DB} \cdot \overset{\large\rightharpoonup}{DC} = \overline{DA}^2 + \overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AC} > 0$，故 $\angle BDC$ 必為銳角。$(3)$ 正確。 - 若 $\angle BAC$ 是鈍角 $\implies \overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AC} < 0$，但若 $\overline{DA}$ 足夠大，其內積仍可能為正，即 $\angle BDC$ 可為銳角。$(4)$ 錯誤。 - 若 $\overline{AB} < \overline{AD}$ 且 $\overline{AC} < \overline{AD}$，由柯西不等式： $$|\overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AC}| \le \overline{AB} \cdot \overline{AC} < \overline{AD} \cdot \overline{AD} = \overline{DA}^2 \implies \overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AC} > -\overline{DA}^2$$ 因此 $\overset{\large\rightharpoonup}{DB} \cdot \overset{\large\rightharpoonup}{DC} = \overline{DA}^2 + \overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AC} > 0$，故 $\angle BDC$ 必為銳角。$(5)$ 正確。 故選 $(3)(5)$。