8.4: Angle Bisectors

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If $\measuredangle ABX \equiv - \measuredangle CBX$, then we say that the line $(BX)$ bisects $\angle ABC$, or the line $(BX)$ is a bisector of $\angle ABC$. If $\measuredangle ABX \equiv \pi - \measuredangle CBX$, then the line $(BX)$ is called the external bisector of $\angle ABC$.

$截屏2021-02-17 下午1.11.35.png$

If $\measuredangle ABA' = \pi$; that is, if $B$ lies between $A$ and $A'$, then the bisector of $\angle ABC$ is the external bisector of $\angle A'BC$ and the other way around.

Note that the bisector and the external bisector are uniquely defined by the angle.

Exercise $\PageIndex{1}$

Show that for any angle, its bisector and external bisector are perpendicular.

Hint

Let $(BX)$ and $(BY)$ be the internal and external bisectors of $\angle ABC$. Then

$2 \cdot \measuredangle XBY \equiv 2 \cdot \measuredangle XBA + 2 \cdot \measuredangle ABY \equiv \measuredangle CBA + \pi + 2 \cdot \measuredangle ABC \equiv \pi + \measuredangle CBC = \pi$

and hence the result.

The bisectors of $\angle ABC$, $\angle BCA$, and $\angle CAB$ of a nondegenerate triangle $ABC$ are called bisectors of the triangle $ABC$ at vertexes $A, B$, and $C$ respectively.

Lemma $\PageIndex{1}$

Let $\triangle ABC$ be a nondegenerate triangle. Assume that the bisector at the vertex $A$ intersects the side $[BC]$ at the point $D$. Then

\[\dfrac{AB}{AC} = \dfrac{DB}{DC}.\]

Proof

$截屏2021-02-17 下午1.18.49.png$

Let $\ell$ be a line passing thru $C$ that is parallel to $(AB)$. Note that $\ell \nparallel (AD)$; set

$E = \ell \cap (AD)$.

Also note that $B$ and $C$ lie on opposite sides of $(AD)$. Therefore, by the transversal property (Theorem 7.3.1),

\[\measuredangle BAD = \measuredangle CED.\]

Further, the angles $ADB$ and $EDC$ are vertical; in particular, by Proposition 2.5.1

$\measuredangle ADB = \measuredangle EDC.$

By the AA similarity condition, $\triangle ABD \sim \triangle ECD$. In particular,

\[\dfrac{AB}{EC} = \dfrac{DB}{DC}.\]

Since $(AD)$ bisects $\angle BAC$, we get that $\measuredangle BAD = \measuredangle DAC$. Together with 8.4.2, it implies that $\measuredangle CEA = \measuredangle EAC$. By Theorem 4.3.1, $\triangle ACE$ is isosceles; that is,

$EC = AC.$

Together with 8.4.3, it implies 8.4.1.

Exercise $\PageIndex{2}$

Formulate and prove an analog of Lemma $\PageIndex{1}$ for the external bisector.

Hint: If $E$ is the point of intersection of $(BC)$ with the external bisector of $\angle BAC$, then $\dfrac{AB}{AC} = \dfrac{EB}{EC}$. It can be proved along the same lines as Lemma $\PageIndex{1}$.

Exercise \(\PageIndex{1}\)

Lemma \(\PageIndex{1}\)

Exercise \(\PageIndex{2}\)