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# 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 $$\angle ABC$$.

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

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 $$\PageIndex{1}$$.