Angle Bisectors and the Incenter

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Learning goals

Measure a point's distance to a side line along a perpendicular
Prove the internal angle-bisector equidistance theorem and its interior-point converse
Construct an angle bisector and justify the equal compass settings
Explain why three internal bisectors meet at the triangle's incenter
Determine the inradius and construct the incircle using a perpendicular foot

Distance goes straight across

The distance from a point to a line is the length of the perpendicular segment from the point to that line. Its endpoint on the line is the perpendicular foot. Any slanted segment to the line is longer, because it is the hypotenuse of a right triangle with the perpendicular as a leg.

For example, if the perpendicular length is 6 and you move 8 along the line from its foot, the slanted length is 10 by Pythagoras. The distance to the line is still 6.

In a triangle, a supporting side line is the whole line containing a side segment. Extend the side when the perpendicular foot falls beyond a vertex. Distance to that line need not equal distance to the side segment: if the foot misses the segment, the nearest point of the segment is an endpoint.

Worked example 1 Choose the perpendicular measurement

Problem. Point $P$ lies off line $\ell$ . Points $X$ and $Y$ lie on $\ell$ , $PX$ is perpendicular to $\ell$ , and the lengths are

\begin{aligned} PX&=9\text{ cm}\\ XY&=12\text{ cm}. \end{aligned}

Find $PY$ and the distance from $P$ to $\ell$ .

The right-angle mark identifies the distance to the line.

Solution. The right triangle has legs 9 and 12.

\begin{aligned} PY^2&=9^2+12^2\\ PY^2&=225\\ PY&=15\text{ cm}. \end{aligned}

The distance from $P$ to $\ell$ is the perpendicular length, $9\text{ cm}$ . The longer slanted segment does not measure that distance.

Check your understanding

A perpendicular from P to a side line has length 7 cm. A slanted segment from P to that line has length 11 cm. What is P's distance to the line?

A 4 cm B 7 cm C 11 cm D 18 cm

Equal angle halves give equal distances

An internal bisector puts a point equally far from the two side lines of an angle.

Take an angle of 60 degrees. A point on its internal bisector makes a 30-degree angle with each side ray. Dropping the two perpendiculars produces right triangles with the same hypotenuse and matching angles. They fit exactly, so their perpendicular legs have equal lengths.

The two half-angles are equal; compare the right triangles sharing OP.

The internal angle-bisector equidistance theorem says that a point on the internal bisector of an angle strictly between zero and 180 degrees has equal perpendicular distances to its two supporting side lines. The vertex itself has distance zero to both lines.

Why an internal angle bisector gives equal distances#

For a concrete case, let the full angle be 68 degrees and let $OP=10$ . If $P$ is on its internal bisector, each half-angle is 34 degrees. The perpendiculars $PX$ and $PY$ form right triangles $OXP$ and $OYP$ . Their right angles match, their 34-degree angles match, and the common side $OP$ matches.

AAS makes the triangles congruent, so $PX=PY$ .

Now let $P$ be any point other than $O$ on the internal bisector of $\angle UOV$ . Each half-angle is acute, so the perpendicular feet $X$ and $Y$ lie on the forward rays $OU$ and $OV$ . In the two right triangles,

\begin{aligned} \angle XOP&\cong\angle POY\\ \angle OXP&\cong\angle OYP. \end{aligned}

Side $OP$ is common. By AAS, the triangles are congruent. Corresponding perpendicular legs give $PX=PY$ . No step depended on the chosen angle or hypotenuse length.

Worked example 2 Use equality of distances

Problem. Point $P$ is on the internal bisector of an angle. Its perpendicular distances to the two side lines are $2x+1$ centimeters and $5x-8$ centimeters. Find $x$ and the common distance.

Solution. Equal distances give one equation.

\begin{aligned} 2x+1&=5x-8\\ 9&=3x\\ x&=3. \end{aligned}

Substitute into each given expression separately.

\begin{aligned} 2(3)+1&=7\\ 5(3)-8&=7. \end{aligned}

Both perpendicular lengths are $7\text{ cm}$ , a positive distance.

Check your understanding

P is on the internal bisector of angle UOV. X and Y are the perpendicular feet on its side lines. Which pair must have equal lengths?

A OU and OV B OP and PX C PU and PV D PX and PY

The converse needs an interior point

Equal perpendicular distances also locate an internal bisector, provided the point is inside the angle.

For instance, an interior point $P$ has two perpendicular lengths of 5 and a vertex distance $OP=13$ . Each right triangle then has its other leg equal to 12. The matching triangles show that $OP$ divides the angle into equal halves.

The interior-point converse states that a point strictly inside an angle between zero and 180 degrees, equidistant from its two supporting side lines, lies on the internal bisector.

Why equal distances locate the internal bisector#

In the 5-12-13 case, the right triangles share hypotenuse 13 and have equal perpendicular legs 5. HL makes them congruent. When the feet are on the forward rays, their equal corresponding vertex angles are precisely the two parts of the original angle.

For a general interior point $P$ , let $X$ and $Y$ be the perpendicular feet on the full lines. Suppose $PX=PY$ . First check the feet, rather than assuming their positions from a drawing.

Neither foot can be $O$ . If one were, its perpendicular length would be $OP$ , while the other perpendicular would be a leg of a nondegenerate right triangle with hypotenuse $OP$ , hence shorter. The two feet cannot both be behind the vertex: that would make both parts of the interior angle at least 90 degrees.

If just one foot were behind the vertex, HL would still make the two right triangles congruent. The acute angle formed with that backward ray would equal the angle formed with the forward ray. The original two angle parts would then be supplementary, making the whole angle 180 degrees. This contradicts the hypothesis. Thus both feet lie on the forward rays.

The right triangles $OXP$ and $OYP$ have common hypotenuse $OP$ and equal legs $PX,PY$ . HL makes them congruent, giving

\angle XOP\cong\angle POY.

Those are the two interior angle parts, so ray $OP$ is the internal bisector.

For two intersecting full lines, external bisectors supply other equal-distance points. A point outside the angle cannot be put on its internal bisector by this converse. The optional Go deeper example shows the distinction on a right angle.

Worked example 3 Check the hypothesis before reversing

Problem. Point $R$ is strictly inside $\angle ABC$ . Its perpendicular distances to lines $BA$ and $BC$ are both $4\text{ cm}$ . If $\angle ABC=74^\circ$ , find $\angle ABR$ and explain which condition selects the internal bisector.

Solution. The equal distances and interior location satisfy the converse. Ray $BR$ is the internal bisector.

\begin{aligned} \angle ABR&=\frac{74^\circ}{2}\\ \angle ABR&=37^\circ. \end{aligned}

Being strictly inside the angle selects the internal ray. Equal distances to full lines alone would allow exterior possibilities.

Check your understanding

Which facts guarantee that OP is the internal bisector of an angle less than 180 degrees?

A P is inside the angle and its perpendicular distances to the side lines are equal B P is outside and its distances to the vertex are equal C P is equally far from two chosen points on the rays D One perpendicular distance from P is known

Construct an angle bisector

Equal compass settings can make equal angle halves without a protractor.

Center a circle arc at vertex $O$ so it meets the two side rays at distinct points $U,V$ . The equal radii give $OU=OV$ .
Keep that compass radius. Draw arcs centered at $U$ and $V$ . Besides $O$ , they meet at a second point $X$ inside the angle. Use that intersection, not $O$ again.
Draw ray $OX$ with a straightedge.

Equal compass radii supply two side pairs; OX supplies the third.

Why the compass construction bisects the angle#

For an initial compass radius of $4\text{ cm}$ , the first arc gives equal lengths $OU,OV$ . The two later arcs give equal lengths $UX,VX$ , also $4\text{ cm}$ . The two triangles share $OX$ , so SSS makes them congruent and their angles at $O$ equal.

With any positive initial radius, the first circle gives $OU=OV$ .

The equal settings in the second pair give $UX=VX$ .

Side $OX$ is common, so triangles $OUX$ and $OVX$ are congruent by SSS. Since $X$ is the intersection inside the angle, the equal corresponding angles at $O$ are its two interior halves. A zero angle or a straight angle is excluded from this construction.

Worked example 4 Justify each compass equality

Problem. An arc centered at $O$ meets the rays of an angle at $U,V$ . Two arcs centered at $U,V$ , drawn with a common radius large enough to cross inside the angle, meet at $X$ . Name the three side equalities that prove ray $OX$ bisects the angle. Must the second common radius equal the first radius?

Solution. The equalities for SSS are

\begin{aligned} OU&=OV\\ UX&=VX\\ OX&=OX. \end{aligned}

The first pair comes from the first arc, the second from equal settings for the two later arcs, and the third from the shared side. SSS gives equal angles at $O$ . The second common radius may differ from the first, provided the arcs cross at a point strictly inside the angle. The two later settings must agree with each other.

Check your understanding

After the first arc marks U and V, which later step preserves the SSS justification?

A Use unrelated radii centered at U and V B Join U to V and call it the bisector C Use equal radii centered at U and V and choose their interior intersection D Choose any point inside the angle

Three bisectors meet, and one circle fits

In any nondegenerate triangle, the three internal angle bisectors meet at one interior point. This point is the incenter.

The distance argument is short: a point on the first bisector has equal distances to two side lines. Being on a second bisector links the third distance to those two. The converse then puts the point on the third bisector.

Why the internal angle bisectors are concurrent#

First try distances of $3\text{ cm}$ . A point $I$ on the bisector at $A$ is 3 centimeters from $AB$ and $AC$ . If it is also on the bisector at $B$ , its distance to $BC$ is 3 centimeters too. If $I$ is inside the triangle, the converse at $C$ puts it on the third internal bisector.

Now establish the point and the general claim. The internal bisector from $A$ meets segment $BC$ at an interior point $K$ . The internal bisector from $B$ meets $AC$ at an interior point $L$ . Points $B$ and $L$ lie on opposite sides of line $AK$ , so segment $BL$ crosses $AK$ at a point $I$ inside the triangle. Thus two internal bisectors really do meet there.

Let $D,E,F$ be the feet from $I$ to lines $BC,CA,AB$ , respectively. The theorem at $A$ gives $IF=IE$ .

The theorem at $B$ gives $IF=ID$ .

Hence $ID=IE$ . Since $I$ lies inside angle $C$ , the converse puts $CI$ on its internal bisector.

The two bisectors link three perpendicular distances; the converse includes the third bisector.

The first two bisector lines are distinct, so they have only one intersection. This proves concurrency, meaning that all three pass through the same point.

The common perpendicular distance is the inradius, written $r$ . The incircle is the circle centered at the incenter with radius $r$ ; it lies inside the triangle and touches each side once.

The feet really are on the side segments. For side $AB$ , the bisector at $A$ puts its foot on ray $AB$ , and the bisector at $B$ puts that same foot on ray $BA$ . The foot is therefore strictly between $A$ and $B$ . The same reasoning applies to the other sides, even when the triangle is obtuse.

To justify the circle, every point of line $AB$ other than $F$ is farther from $I$ than $IF$ : Pythagoras makes the slanted segment longer. Thus the circle meets that side line exactly at $F$ . The other two lines behave the same way. A point beyond a side line would be farther than $r$ , since a segment from $I$ to it must cross that line first. The whole circle therefore stays in the triangle.

To construct the incircle, construct any two internal bisectors and label their intersection $I$ . Construct a perpendicular from $I$ to line $AB$ and label its foot $F$ . Set the compass to $IF$ and draw the circle centered at $I$ . Using $IA$ instead would draw a circle through a vertex, not the incircle.

The circle reaches each side at its perpendicular foot, with equal radii.

Use Back and Next to build the internal bisectors. Drag A or vary either base angle; BC stays fixed at 6 units. Follow the square marks to the three perpendicular feet, then compare their distances from I. These measured observations support the geometric argument above.

One interior center has three equal perpendicular distances

Construction: complete construction

Scroll sideways to see the whole diagram.

Angle B Angle C Construction Try a case

Angle B = 55°. Angle C = 65°. Angle A = 60°. The triangle is acute. The three internal angle bisectors meet at I. The perpendicular distance from I to BC is approximately 1.72 units. The perpendicular distance from I to CA is approximately 1.72 units. The perpendicular distance from I to AB is approximately 1.72 units. I is inside the triangle. The incircle touches all three side segments.

The incircle uses the perpendicular distance from I to a side line as its radius. Its feet remain on the side segments as the triangle changes.

Worked example 5 Find the compass radius from a right triangle

Problem. $I$ is the incenter of triangle $ABC$ , and $F$ is the perpendicular foot on side $AB$ . The lengths are

\begin{aligned} AI&=13\text{ cm}\\ AF&=12\text{ cm}. \end{aligned}

Find the inradius and the perpendicular distances from $I$ to the other two sides. State the center and radius to use when drawing the incircle.

A detail near A shows the right triangle needed for the radius.

Solution. In right triangle $AFI$ , $AI$ is the hypotenuse.

\begin{aligned} IF^2&=13^2-12^2\\ IF^2&=25\\ IF&=5\text{ cm}. \end{aligned}

The inradius is $5\text{ cm}$ , and each other perpendicular distance is also $5\text{ cm}$ . Center the compass at $I$ with radius $IF$ , which is $5\text{ cm}$ .

Check your understanding

Two internal angle bisectors of triangle ABC meet at I. What completes a compass-and-straightedge construction of the incircle?

A Use a vertex as center and radius IA B Use I as center and radius IA C Draw a circle of any radius centered at I D Construct a perpendicular IF to side AB, then use center I and radius IF

Angle Bisectors and the Incenter

Learning goals

Distance goes straight across

Equal angle halves give equal distances

The converse needs an interior point

Construct an angle bisector

Three bisectors meet, and one circle fits

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Multiple Choice Questions (MCQ)

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