問題の解答
検索用コード（LaTeX）
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% 例題I4.3.6：角の二等分線の長さ（One More）★★
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$\triangle\mathrm{ABC}$において，$\mathrm{AB}=4,\mathrm{BC}=8,\mathrm{CA}=6$とする．$\angle\mathrm{A}$の二等分線が辺BCと交わる点を$\mathrm{D}$，$\angle\mathrm{B}$の二等分線が線分ADと交わる点をIとする．このとき，次の値を求めよ．
(1) 線分ADの長さ
(2) 線分AIの長さ
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% 例題I4.3.6の解答（One More）★★
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(1) $\triangle\mathrm{ABC}$において，線分ADは$\angle\mathrm{A}$の二等分線であるから，
\[
\mathrm{BD}:\mathrm{DC}=\mathrm{AB}:\mathrm{AC}=4:6=2:3
\]
$\mathrm{BC}=8$より，
\[
\mathrm{BD}=\frac{2}{2+3}\mathrm{BC}=\frac{2}{5}\cdot 8=\frac{16}{5}
\]
$\triangle\mathrm{ABC}$において，余弦定理より，
\[
\cos B=\frac{\mathrm{AB}^2+\mathrm{BC}^2-\mathrm{CA}^2}{2\mathrm{AB}\cdot\mathrm{BC}}=\frac{4^2+8^2-6^2}{2\cdot 4\cdot 8}=\frac{11}{16}
\]
$\triangle\mathrm{ABD}$において，余弦定理より，
\[
\mathrm{AD}^2=\mathrm{AB}^2+\mathrm{BD}^2-2\mathrm{AB}\cdot\mathrm{BD}\cdot\cos B=4^2+(\frac{16}{5})^2-2\cdot 4\cdot\frac{16}{5}\cdot\frac{11}{16}=\frac{216}{25}
\]
よって，$\mathrm{AD}>0$より，$\mathrm{AD}=\frac{6\sqrt{6}}{5}$
(2) $\triangle\mathrm{ABD}$において，線分BIは$\angle\mathrm{B}$の二等分線であるから，
\[
\mathrm{AI}:\mathrm{ID}=\mathrm{AB}:\mathrm{BD}=4:\frac{16}{5}=5:4
\]
よって，$\mathrm{AI}=\frac{5}{5+4}\mathrm{AD}=\frac{5}{9}\times\frac{6\sqrt{6}}{5}=\frac{2\sqrt{6}}{3}$
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% 問題I4.3.6：角の二等分線の長さ（One More）★★
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$\triangle\mathrm{ABC}$において，$\mathrm{AB}=4,\mathrm{BC}=6,\mathrm{CA}=5$とする．$\angle\mathrm{A}$の二等分線が辺BCと交わる点を$\mathrm{D}$，$\angle\mathrm{B}$の二等分線が線分ADと交わる点をIとする．このとき，次の値を求めよ．
(1) 線分ADの長さ
(2) 線分AIの長さ
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% 問題I4.3.6の解答（One More）★★
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(1) $\triangle\mathrm{ABC}$において，線分ADは$\angle\mathrm{A}$の二等分線であるから，
\[
\mathrm{BD}:\mathrm{DC}=\mathrm{AB}:\mathrm{AC}=4:5
\]
$\mathrm{BC}=6$より，
\[
\mathrm{BD}=\frac{4}{4+5}\mathrm{BC}=\frac{4}{9}\cdot 6=\frac{8}{3}
\]
$\triangle\mathrm{ABC}$において，余弦定理より，
\[
\cos B=\frac{\mathrm{AB}^2+\mathrm{BC}^2-\mathrm{CA}^2}{2\mathrm{AB}\cdot\mathrm{BC}}=\frac{4^2+6^2-5^2}{2\cdot 4\cdot 6}=\frac{9}{16}
\]
$\triangle\mathrm{ABD}$において，余弦定理より，
\[
\mathrm{AD}^2=\mathrm{AB}^2+\mathrm{BD}^2-2\mathrm{AB}\cdot\mathrm{BD}\cdot\cos B=4^2+(\frac{8}{3})^2-2\cdot 4\cdot\frac{8}{3}\cdot\frac{9}{16}=\frac{100}{9}
\]
よって，$\mathrm{AD}>0$より，
\[
\mathrm{AD}=\sqrt{\frac{100}{9}}=\frac{10}{3}
\]
(2) $\triangle\mathrm{ABD}$において，線分BIは$\angle\mathrm{B}$の二等分線であるから，
\[
\mathrm{AI}:\mathrm{ID}=\mathrm{AB}:\mathrm{BD}=4:\frac{8}{3}=3:2
\]
よって，$\mathrm{AI}=\frac{3}{3+2}\mathrm{AD}=\frac{3}{5}\times\frac{10}{3}=2$
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