角平分線定理

${\displaystyle \angle \alpha =\angle \beta }$${\displaystyle \;\Leftrightarrow \;}$${\displaystyle AE=AC}$${\displaystyle \;\Leftrightarrow \;}$${\displaystyle {\tfrac {DB}{DC}}={\tfrac {AB}{AE}}={\tfrac {AB}{AC}}}$

在 ${\displaystyle \triangle ABC}$ 中，在 ${\displaystyle BC}$ 边上任取一点 ${\displaystyle D}$ 。过点 ${\displaystyle C}$ 做 ${\displaystyle AD}$ 的平行线，与 ${\displaystyle {\overline {BA}}}$ 的延长线相交于点 ${\displaystyle E}$ 。

${\displaystyle \angle BAD=\angle AEC}$

${\displaystyle \angle CAD=\angle ACE}$

${\displaystyle \triangle DBA}$ ${\displaystyle {\color {Blue}\sim }}$ ${\displaystyle \triangle CBE}$${\displaystyle \;\Rightarrow \;}$${\displaystyle {DB \over DC}={AB \over AE}}$

證内角平分线定理

${\displaystyle \angle BAD=\angle CAD}$${\displaystyle \;\Rightarrow \;}$${\displaystyle \angle AEC=\angle ACE}$${\displaystyle \;\Rightarrow \;}$${\displaystyle \triangle ACE}$ 为等腰三角形${\displaystyle \;\Rightarrow \;}$${\displaystyle AC=AE}$${\displaystyle \;\Rightarrow \;}$${\displaystyle {DB \over DC}={AB \over AE}={AB \over AC}}$

證内角平分线逆定理

${\displaystyle AC={AB\cdot DC \over DB}=AE}$${\displaystyle \;\Rightarrow \;}$${\displaystyle \triangle ACE}$ 为等腰三角形${\displaystyle \;\Rightarrow \;}$${\displaystyle \angle AEC=\angle ACE}$${\displaystyle \;\Rightarrow \;}$${\displaystyle \angle BAD=\angle CAD}$

${\displaystyle \angle \alpha =\angle \beta }$${\displaystyle \;\Leftrightarrow \;}$${\displaystyle AE=AC}$${\displaystyle \;\Leftrightarrow \;}$${\displaystyle {\tfrac {DB}{DC}}={\tfrac {AB}{AE}}={\tfrac {AB}{AC}}}$

在 ${\displaystyle \triangle ABC}$ 中，令 ${\displaystyle AB>AC}$ 。在 ${\displaystyle {\overline {BC}}}$ 的延长线上取一点 ${\displaystyle D}$ 。过点 ${\displaystyle C}$ 做 ${\displaystyle AD}$ 的平行线，与 ${\displaystyle BA}$ 边相交于点 ${\displaystyle E}$ 。在 ${\displaystyle BA}$ 的延长线上任取一点 ${\displaystyle F}$ 。

${\displaystyle \angle FAD=\angle AEC}$

${\displaystyle \angle CAD=\angle ACE}$

${\displaystyle \triangle DBA\sim \triangle CBE}$${\displaystyle \;\Rightarrow \;}$${\displaystyle {DB \over DC}={AB \over AE}}$

證外角平分线定理

易证得，三角形外角平分线与对边直线的交点，必定落在较短的邻边的一侧。

${\displaystyle \angle FAD=\angle CAD}$${\displaystyle \;\Rightarrow \;}$${\displaystyle \angle AEC=\angle ACE}$${\displaystyle \;\Rightarrow \;}$${\displaystyle \triangle ACE}$ 为等腰三角形${\displaystyle \;\Rightarrow \;}$${\displaystyle AC=AE}$${\displaystyle \;\Rightarrow \;}$${\displaystyle {DB \over DC}={AB \over AE}={AB \over AC}}$

證外角平分线逆定理

易证得，三角形一边所在直线上符合要求的点，必定落在较短的邻边的一侧。

${\displaystyle AC={AB\cdot DC \over DB}=AE}$${\displaystyle \;\Rightarrow \;}$${\displaystyle \triangle ACE}$ 为等腰三角形${\displaystyle \;\Rightarrow \;}$${\displaystyle \angle AEC=\angle ACE}$${\displaystyle \;\Rightarrow \;}$${\displaystyle \angle FAD=\angle CAD}$