【数学A】例題3.1.1：角の二等分線と比（One More）★★

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% 例題A3.1.1：角の二等分線と比（One More）★★
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$\mathrm{AB}=6,\mathrm{BC}=5,\mathrm{CA}=4$である$\triangle\mathrm{ABC}$において，$\angle\mathrm{A}$およびその外角の二等分線が辺BCまたはその延長と交わる点を，それぞれD，Eとする．このとき，線分DEの長さを求めよ．

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% 例題A3.1.1の解答（One More）★★
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ADは$\angle\mathrm{A}$の二等分線であるから，

\[
\mathrm{BD}:\mathrm{DC}=\mathrm{AB}:\mathrm{AC}
\]

したがって，$\mathrm{BD}:\mathrm{DC}=6:4=3:2$

ゆえに，$\mathrm{DC}:\mathrm{BC}=2:5$であるから，

\[
\mathrm{DC}=\frac{2}{5}\mathrm{BC}=\frac{2}{5}\cdot 5=2
\]

また，AEは$\angle\mathrm{A}$の外角の二等分線であるから，

\[
\mathrm{BE}:\mathrm{EC}=\mathrm{AB}:\mathrm{AC}
\]

したがって，$\mathrm{BE}:\mathrm{EC}=6:4=3:2$

ゆえに，$\mathrm{BC}:\mathrm{EC}=1:2$であるから，

\[
\mathrm{EC}=2\mathrm{BC}=2\cdot 5=10
\]

よって，$\mathrm{DE}=\mathrm{DC}+\mathrm{EC}=2+10=12$

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% 問題A3.1.1：角の二等分線と比（One More）★★
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$\mathrm{AB}=8,\mathrm{BC}=7,\mathrm{CA}=6$である$\triangle\mathrm{ABC}$において，$\angle\mathrm{A}$およびその外角の二等分線が辺BCまたはその延長と交わる点を，それぞれD，Eとする．このとき，線分DEの長さを求めよ．

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% 問題A3.1.1の解答（One More）★★
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ADは$\angle\mathrm{A}$の二等分線であるから，

\[
\mathrm{BD}:\mathrm{DC}=\mathrm{AB}:\mathrm{AC}
\]

したがって，$\mathrm{BD}:\mathrm{DC}=8:6=4:3$

ゆえに，$\mathrm{DC}:\mathrm{BC}=3:7$であるから，

\[
\mathrm{DC}=\frac{3}{7}\mathrm{BC}=\frac{3}{7}\cdot 7=3
\]

また，AEは$\angle\mathrm{A}$の外角の二等分線であるから，

\[
\mathrm{BE}:\mathrm{EC}=\mathrm{AB}:\mathrm{AC}
\]

したがって，$\mathrm{BE}:\mathrm{EC}=8:6=4:3$

ゆえに，$\mathrm{BC}:\mathrm{EC}=1:3$であるから，

\[
\mathrm{EC}=3\mathrm{BC}=3\cdot 7=21
\]

よって，$\mathrm{DE}=\mathrm{DC}+\mathrm{EC}=3+21=24$

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% 例題A3.1.1の側注：角の二等分線と比（One More）★★
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% 側注1
$\mathrm{BC}=5$であることを用いて，$\mathrm{BD}:\mathrm{DC}=\mathrm{AB}:\mathrm{AC}$，すなわち，
\[
(5-\mathrm{DC}):\mathrm{DC}=6:4
\]
より，$\mathrm{DC}$の長さを求めてもよい．
\[
6\mathrm{DC}=4(5-\mathrm{DC})
\]
であるから，$\mathrm{DC}=2$となる．

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% 問題A3.1.1の側注：角の二等分線と比（One More）★★
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% 側注1
$\mathrm{BC}={7}$であることを用いて，$\mathrm{BD}:\mathrm{DC}=\mathrm{AB}:\mathrm{AC}$，すなわち，
\[
({7}-\mathrm{DC}):\mathrm{DC}=8:6
\]
より，$\mathrm{DC}$の長さを求めてもよい．
\[
8\mathrm{DC}=6(7-\mathrm{DC})
\]
であるから，$\mathrm{DC}=3$となる．

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% 例題A3.1.1の考え方・補足など：角の二等分線と比（One More）★★
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% 考え方
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線分$\mathrm{DC},\mathrm{EC}$の長さを求めることを考える．

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$\triangle\mathrm{ABC}$の辺$\mathrm{BC}$を内分する点$\mathrm{D}$について，

\[
\angle\mathrm{BAD}=\angle\mathrm{CAD}\Longleftrightarrow\mathrm{AB:AC=BD:DC}
\]

$\triangle\mathrm{ABC}$の辺$\mathrm{BC}$を外分する点$\mathrm{E}$について，

\[
\angle\mathrm{CAE}=\angle\mathrm{FAE}\Longleftrightarrow\mathrm{AB:AC=BE:EC}
\]