Angle Bisector Theorem

The angle bisector theorem is concerned with the relative lengths of the two segments that a triangle's side is divided into by a line that bisects the opposite angle. It equates their relative lengths to the relative lengths of the other two sides of the triangle.

To bisect an angle means to cut it into two equal parts or angles. Say that we wanted to bisect a 50-degree angle, then we would divide it into two 25-degree angles.

Angle Bisector Theorem

In the figure above, let $x+y=c\big(=\lvert\overline{BA}\rvert\big),$ and let $\lvert\overline{CD}\rvert=e$ be the length of the bisector of angle $C$.

In its simplest form, the angle bisector theorem states that

$\begin{array}{c}&\frac{a}{x}=\frac{b}{y} &\text{ or } &ay=bx.\qquad (1) \end{array}$

The angle bisector theorem also states that $e,$ the length of the angle bisector $\overline{CD},$ satisfies

$e^2=ab-xy. \qquad (2)$

Proof of (1):

Applying the sine rule on $\triangle BCD$ and $\triangle ACD$ gives

$\dfrac{a}{\sin\angle BDC}=\dfrac{x}{\sin\angle BCD}, ~~~~~\dfrac{b}{\sin\angle ADC}=\dfrac{y}{\sin\angle ACD}.$

Using the equalities $\sin\angle ADC=\sin\left(\pi-\angle BDC\right)=\sin\angle BDC$ and $\angle BCD=\angle ACD$ $($since $CD$ is the angle bisector$),$ we get

$\dfrac{a}{x}=\dfrac{\sin\angle BDC}{\sin\angle BCD}=\dfrac{\sin\angle ADC}{\sin\angle ACD}=\dfrac{b}{y},$

which is what we want. $_\square$

Proof of (2):

Stewart's theorem states that (remember $x+y=c$)

$xy(x+y)+e^2(x+y)=a^2y+b^2x.$

Rearranging gives $e^2=\frac{a^2y+b^2x}{x+y}-xy.$ Then using the angle bisector theorem $x=\frac{ay}{b}$ and $y=\frac{bx}{a},$ we have

$\begin{aligned} e^2 &=\dfrac{a^2y+b^2x}{\frac{ay}{b}+\frac{bx}{a}}-xy\\ &=\dfrac{a^2y+b^2x}{\frac{a^2y+b^2x}{ab}}-xy\\ &=ab-xy. \ _\square \end{aligned}$

Another proof of (1):

Let us consider $\triangle ABC$ and let $AE$ be the angular bisector of $\angle A$.

Now extend $AE$ to $D$ such that $CD$ be parallel to $AB$.

Then $\angle ABE$ = $\angle DCE$ and $\angle BAE$ = $\angle CDE,$ which implies

$\triangle ABE\sim \triangle DCE.$

So,

$\dfrac{c}{x} = \dfrac{a}{y}.$

Since $\angle BAE$ = $\angle CDE$ = $\angle CAE,$ by isosceles property $a = b,$ which implies

$\dfrac{c}{x} = \dfrac{b}{y}. \ _\square$

In $\triangle{ABC}$, $\lvert\overline{AB}\rvert=10, \lvert\overline{BC}\rvert=8, \lvert\overline{AC}\rvert=12$. Let $D$ be a point on side $\overline{AB}$ such that $\overline{CD}$ bisects $\angle C$. Then what is the length of $\overline{AD}?$

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Let $\lvert\overline{AB}\rvert=c, \lvert\overline{BC}\rvert=a, \lvert\overline{AC}\rvert=b, \lvert\overline{AD}\rvert=y, \lvert\overline{BD}\rvert=x$, then we are now looking for $y.$

Using the angle bisector theorem, $\frac{y}{b}=\frac{x}{a}\implies \frac{y}{12}=\frac{x}{8}. \qquad (1)$

Since $x+y=c$ or $x=c-y,$ plugging this into $(1)$, we have

$\begin{aligned} \frac{y}{12}&=\frac{10-y}{8}\\\\ \Rightarrow y&=6. \ _\square \end{aligned}$

This next example is the same as the previous, but we are instead solving for the length $e$ of the angle bisector $\overline{CD}$.

In $\triangle{ABC}$, $\lvert\overline{AB}\rvert=10, \lvert\overline{BC}\rvert=8, \lvert\overline{AC}\rvert=12$. Let $D$ be a point on side $\overline{AB}$ such that $\overline{CD}$ bisects $\angle C$, then what is the length of $\overline{CD}?$

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In our previous example, we already found $\lvert\overline{AD}\rvert=6$ and $\lvert\overline{BD}\rvert=4$.

Thus,

$\begin{aligned} e &=\sqrt{ab-xy}\\ &= \sqrt{(8)(12)-(4)(6)}\\ &=6\sqrt{2}. \ _\square \end{aligned}$

We are given a triangle with the following property: one of its angles is quadrisected (divided into four equal angles) by the height, the angle bisector, and the median from that vertex.

Find the measure of the quadrisected angle.

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The base is partitioned into four segments in the ratio $x : x : y : 2x +y$.

Suppose the length of the left-hand side of the triangle is $1$.
Then the length of the angle bisector is also $1$.
Applying the angle bisector theorem to the large triangle, we see that the length of the right-hand side is

$\dfrac{2x+2y}{2x}=1+\dfrac{y}{x}.$

But if we apply the angle bisector theorem to the left half of the triangle, we obtain $\frac{2x+y}{y}= 1+\frac{2x}{y}$ for the same length. Therefore,

$\dfrac{y}{x}=\dfrac{2x}{y} \implies x:y=1:\sqrt2.$

Now apply the angle bisector theorem a third time to the right triangle formed by the altitude and the median. The segments in the base are in the ratio $x:y=1:\sqrt2$, so the altitude and the median form the same ratio. As this is a right triangle, it must be a 45$^\circ$-45$^\circ$-90$^\circ$ triangle.

So the quadrisected angle is right. $_\square$

Some additional problems are as follows:

1. Find the length of the internal bisector of the right angle in the right triangle with side lengths $3,4,5$.

2. Let $ABC$ be a triangle with angle bisector $AD$ with $D$ on side $BC$. If $\lvert\overline{BD}\rvert = 2, \lvert\overline{CD}\rvert = 5$, and $\lvert\overline{AB} +\lvert\overline{AC}\rvert = 10$, what are $\lvert\overline{AB}\rvert$ and $\lvert\overline{AC}\rvert?$