解答・解説[3](3)

\begin{eqnarray*} \overrightarrow{\mathrm{AD}} &=& \overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BD}} \end{eqnarray*}
\(点\mathrm Bの直線\mathrm{AP}に関する対称点が\mathrm Dだから\) \begin{eqnarray*} \overrightarrow{\mathrm{AD}} &=& \overrightarrow{\mathrm{AB}} + 2\overrightarrow{\mathrm{BH}} \\[3mm] &=& \overrightarrow{\mathrm{AB}} +2(\overrightarrow{\mathrm{AH}} - \overrightarrow{\mathrm{AB}}) \\[3mm] &=& \overrightarrow{\mathrm{AB}} + 2\overrightarrow{\mathrm{AH}} -2\overrightarrow{\mathrm{AB}} \\[3mm] &=& -\overrightarrow{\mathrm{AB}} + 2\overrightarrow{\mathrm{AH}} \\[3mm] &=& -\overrightarrow{\mathrm{AB}} + 2・\frac{6}{7}\overrightarrow{\mathrm{AP}} \\[3mm] &=& -\overrightarrow{\mathrm{AB}} + \frac{12}{7}\left(\frac{3\overrightarrow{\mathrm{AH}}+\overrightarrow{\mathrm{AC}}}{4}\right) \\[3mm] &=& -\overrightarrow{\mathrm{AB}} + \frac{9}{7}\overrightarrow{\mathrm{AB}} + \frac{3}{7}\overrightarrow{\mathrm{AC}} \\[3mm] &=& \frac{2}{7}\overrightarrow{\mathrm{AB}} + \frac{3}{7}\overrightarrow{\mathrm{AC}} \\[3mm] \end{eqnarray*}