Geometry

Angle bisectors of a triangle

A point sits inside an angle. When is it exactly as far from one side as from the other, and what does that tell you about the three angles of a triangle?

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Distance from a point to a lineGeometry Congruent angles, angle addition and bisectorsMath 7

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IncenterGeometry Triangle proportionality theoremGeometry Triangle angle bisector theoremGeometry

Where you are 4 in Special segments in a triangle

Geometry11

Triangles18

Special segments in a triangle4

Medians and altitudes

Angle bisectors of a triangle

Midsegments of triangles and trapezoids Perpendicular bisectors of a triangle

Why this exists

Two straight fences meet at the corner of a field and make an angle. A sprinkler is to stand inside the corner, exactly as far from one fence as from the other. Where can it go?

You could try a spot, measure to each fence, and move it a little. That finds one good spot at a time and never tells you whether you have found them all. The answer in this lesson finds every one at once: the good spots are exactly the points of the angle bisector, the ray that cuts the corner into two equal angles. Every point on it works, and no other point inside the corner does.

Now make the field a triangle, with three fences. Is there a spot the same distance from all three? Each pair of fences has its own bisector, and the surprise is that the three bisectors of a triangle always cross at a single point. That point is the same distance from every side, so a circle drawn around it can touch all three sides at once. The rest of that story belongs to the Incenter; this lesson proves the facts it stands on.

Definition

The distance from a point to a line is the length of the perpendicular segment from the point to the line, as in Distance from a point to a line. The end of that segment on the line is the foot of the perpendicular.

The distance from a point to a side of an angle is its distance to the line containing that side. A point P is equidistant from the sides of \angle ABC when its perpendicular segments to line BA and to line BC have the same length.

A point P is inside \angle ABC (in its interior) when it lies on the same side of line BA as C, and on the same side of line BC as A.

In Figure 1, \overrightarrow{BD} bisects \angle ABC, so m\angle ABD = m\angle DBC. The point P is on the bisector, \overline{PF} \perp \overline{BA} and \overline{PG} \perp \overline{BC}. The ticks say PF = PG.

Figure 1A point on the bisector. Its two perpendiculars to the sides are equal.

In a triangle, the angle bisector from a vertex is the segment that bisects the angle at that vertex and runs to the opposite side. In \triangle ABC the bisector from A starts at A, bisects \angle BAC and ends on \overline{BC}. Every triangle has three of them, one from each vertex.

Intuition

Fold the page along the bisector. From Congruent angles, angle addition and bisectors, the fold lays ray BA exactly on ray BC, because the two halves of the angle are equal. The point P is on the crease, so it does not move.

Now watch the segment \overline{PF}. It meets BA at a right angle, and folding keeps angles and lengths, so after the fold it is a segment from P meeting BC at a right angle. From one point there is only one perpendicular to a line, so the folded copy of \overline{PF} is exactly \overline{PG}. They match, so PF = PG.

Off the crease the fold does not help, and Figure 2 shows what happens instead. The point Q is inside the angle but below the bisector, leaning toward side BC. It is closer to BC than to BA. Slide Q upward and the two distances trade places: at first QG is the shorter one, and once Q crosses the bisector QF is. The bisector is the dividing line where the two are equal.

Figure 2Off the bisector, the point leans toward one side and the two distances differ.

Results

The angle bisector theorem

If a point lies on the bisector of an angle, then it is the same distance from the two sides of the angle.

In the notation of Figure 1: if \overrightarrow{BD} bisects \angle ABC and P is on \overrightarrow{BD}, then PF = PG. The proof also gives BF = BG.

Its converse

If a point inside an angle is the same distance from the two sides, then it lies on the bisector of the angle.

The two together

Inside an angle, a point is equidistant from the two sides if and only if it lies on the bisector. So the bisector is the locus of points inside the angle that are equidistant from its sides: it contains all of them and nothing else.

The three bisectors of a triangle

The three angle bisectors of any triangle meet at one point, inside the triangle, and that point is the same distance from all three sides. It is called the Incenter and is usually named I.

Figure 3The three bisectors of a triangle meet at one point, the same distance from all three sides.

Why it is true

A foot lemma

Both proofs need to know where the foot of a perpendicular lands. Suppose \angle PBA is acute. Then the foot F of the perpendicular from P to line BA lies on ray BA itself, not at B and not behind it. If F were B, then \angle PBA would be a right angle. If F were on the opposite ray, then \angle PBF would be the supplement of an acute angle, so obtuse, and triangle PBF would hold an obtuse angle and a right angle, which add to more than 180^\circ (the triangle angle sum). Neither can happen.

The theorem

Let \overrightarrow{BD} bisect \angle ABC, where m\angle ABC < 180^\circ, and let P \ne B be on it. Each half of the angle is less than 90^\circ, so by the lemma the feet F and G land on the sides \overrightarrow{BA} and \overrightarrow{BC}.

Statements	Reasons
\angle PBF \cong \angle PBG	\overrightarrow{BD} bisects \angle ABC
\angle PFB \cong \angle PGB	both are right angles
\overline{BP} \cong \overline{BP}	the same segment (reflexive property)
\triangle PBF \cong \triangle PBG	AAS (SSS, SAS, ASA, AAS)
PF = PG	corresponding parts of congruent triangles
BF = BG	corresponding parts of congruent triangles

If P = B both distances are 0, so the theorem holds there too.

The converse

Let P be inside \angle ABC with PF = PG, where F and G are the feet on lines BA and BC.

When both feet land on the sides. Triangles PBF and PBG have right angles at F and G, share the hypotenuse \overline{BP}, and have the equal legs PF = PG. By hypotenuse leg congruence they are congruent, so \angle PBF \cong \angle PBG, which says m\angle PBA = m\angle PBC. Because P is inside the angle, the angle addition postulate gives m\angle PBA + m\angle PBC = m\angle ABC, so the two equal parts are halves and \overrightarrow{BP} is the bisector.

When a foot misses its side. This can only happen when the angle is obtuse, and it leads to a contradiction. Say F is not on ray BA.

If F = B, then PB = PF = PG. But PG is a leg of a right triangle with hypotenuse PB, so PG < PB unless G = B as well. Then both sides are perpendicular to \overline{BP} at B, which makes \angle ABC a straight angle.
If F is on the ray opposite BA, then \angle PBA is obtuse. Since m\angle PBC = m\angle ABC - m\angle PBA, which is less than 180^\circ - 90^\circ, the angle \angle PBC is acute and, by the lemma, G lands on ray BC. Hypotenuse leg congruence again gives \angle PBF \cong \angle PBG. But \angle PBF and \angle PBA form a linear pair, so m\angle PBA + m\angle PBC = 180^\circ, and again \angle ABC is a straight angle.

Either way the angle would be straight, which it is not. So both feet land on the sides, and the first case applies.

The three bisectors meet

In \triangle ABC, the bisector from A runs from A to a point D on \overline{BC}, and the bisector from B runs from B across the triangle to \overline{AC}. It must cross the segment \overline{AD}, which cuts the triangle in two with B on one side and the far end of the bisector from B on the other. Call the crossing point I. It is inside the triangle.

I is on the bisector of \angle A, so by the theorem it is as far from AB as from AC.
I is on the bisector of \angle B, so it is as far from AB as from BC.
Both distances equal the distance to AB, so I is as far from AC as from BC.

Since I is inside the triangle, it is inside \angle C, and the converse puts it on the bisector of \angle C. So all three bisectors pass through I, and I is the same distance from all three sides.

Worked examples

Example 1

Lengths on the bisector

In Figure 4, \overrightarrow{BD} bisects \angle ABC and P is on it. F and G are the feet of the perpendiculars from P to \overrightarrow{BA} and \overrightarrow{BC}, with PF = 7 and BF = 24. Find PG, BG and BP.

Figure 4Example 1. The perpendicular and the piece of the side it cuts off.

Solution

P is on the bisector, so by the angle bisector theorem it is the same distance from both sides: PG = PF, that is, PG = 7.

The proof of the theorem showed \triangle PBF \cong \triangle PBG, and \overline{BF} and \overline{BG} are corresponding sides, so BG = 24.

\triangle PBF has a right angle at F, so the The Pythagorean theorem gives the hypotenuse:

BP = \sqrt{7^2 + 24^2} = \sqrt{625} = 25

Check. Triangle PBG is the same 7, 24, 25 triangle, and 7^2 + 24^2 = 49 + 576 = 625.

Example 2

Two distances as expressions

In Figure 5, P is on the bisector \overrightarrow{BD} of \angle ABC. The perpendicular from P to BA has length 3x + 5 and the perpendicular from P to BC has length 5x - 9. Find x and the distance from P to each side.

Figure 5Example 2. The two distances written as expressions in x.

Solution

A point on the bisector is equidistant from the sides, so the two lengths are equal:

3x + 5 = 5x - 9

Subtract 3x from both sides and add 9 to both: 14 = 2x, so x = 7.

The distance is 3(7) + 5 = 26.

Check. The other expression gives 5(7) - 9 = 26 as well, and 26 is positive, as a length must be.

Example 3

Equal distances make a bisector

In Figure 6, P is inside \angle ABC, \overline{PF} \perp \overline{BA}, \overline{PG} \perp \overline{BC} and PF = PG. The angle marked 1 is \angle ABP, with m\angle ABP = (4y - 6)^\circ, and the angle marked 2 is \angle PBC, with m\angle PBC = (2y + 18)^\circ. Find y and m\angle ABC.

Figure 6Example 3. Equal distances are given. Is the ray a bisector?

Solution

P is inside the angle and equidistant from its sides, so by the converse \overrightarrow{BP} bisects \angle ABC. Its two parts are equal:

4y - 6 = 2y + 18

Subtract 2y and add 6: 2y = 24, so y = 12. Each part is 4(12) - 6 = 42^\circ, and by angle addition

m\angle ABC = 42^\circ + 42^\circ = 84^\circ

Check. 2(12) + 18 = 42, so the second part agrees.

Example 4

What the converse needs

In each case P is inside \angle ABC, F is on \overrightarrow{BA} and G is on \overrightarrow{BC}. Decide whether \overrightarrow{BP} must bisect the angle.

(a) \overline{PF} \perp \overline{BA}, \overline{PG} \perp \overline{BC} and PF = PG = 6.

(b) PF = PG = 6, with no right angles given.

Solution

(a) Yes. These are the perpendicular distances to the two sides, they are equal, and P is inside the angle. That is exactly what the converse asks for.

(b) Not necessarily. A slanted segment is longer than the perpendicular from the same point, so equal slanted segments say nothing about the distances. For instance, put P inside the angle, far from B, at distance 2 from BC and distance 5 from BA. A circle of radius 6 about P crosses both sides, so segments of length 6 from P to each side exist, yet P is not equidistant from the sides and so, by the theorem, not on the bisector.

(c) Yes, though not by this lesson’s theorem. Triangles BFP and BGP have BF = BG, FP = GP and the shared side \overline{BP}, so they are congruent by SSS. Then \angle FBP \cong \angle GBP, and \overrightarrow{BP} is the bisector. This is the reasoning of the fold in Congruent angles, angle addition and bisectors, not a fact about distances.

Example 5

Two bisectors of a triangle settle the third

In \triangle ABC of Figure 7, m\angle A = 64^\circ and m\angle B = 48^\circ. The bisectors of \angle A and \angle B meet at I, and the perpendicular from I to \overline{AB} has length 4. Find the distances from I to \overline{BC} and to \overline{CA}, and find m\angle ICB and m\angle AIB.

Figure 7Example 5. Two bisectors are drawn; the third must pass through the same point.

Solution

I is on the bisector of \angle B, so it is as far from BC as from AB: the distance to \overline{BC} is 4. I is on the bisector of \angle A, so it is as far from CA as from AB: the distance to \overline{CA} is also 4.

The three bisectors of a triangle meet at one point, so \overline{CI} bisects \angle C. By the triangle angle sum,

m\angle C = 180^\circ - 64^\circ - 48^\circ = 68^\circ

so m\angle ICB = \tfrac{1}{2}(68^\circ) = 34^\circ.

In \triangle AIB the angle at A is half of 64^\circ, which is 32^\circ, and the angle at B is half of 48^\circ, which is 24^\circ. So

m\angle AIB = 180^\circ - 32^\circ - 24^\circ = 124^\circ

Check. 90^\circ + \tfrac{1}{2}(68^\circ) = 124^\circ too. Exercise 21 shows that this shortcut always works.

Example 6

Finding a bisector on the coordinate plane

An angle has its vertex at the origin O. One side is the positive x-axis and the other is the ray from O through (3, 4). Show that P(6, 3) is the same distance from both sides, and decide whether \overrightarrow{OP} bisects the angle.

Solution

The perpendicular from P to the x-axis is vertical, with foot (6, 0). That foot is on the positive x-axis, and the distance is 3.

The other side lies on the line through the origin with slope \tfrac{4}{3}, the line y = \tfrac{4}{3}x. A perpendicular to it has slope -\tfrac{3}{4} (perpendicular slopes multiply to -1), so the perpendicular through P is y = 3 - \tfrac{3}{4}(x - 6). The foot is where the two lines meet:

\tfrac{4}{3}x = 3 - \tfrac{3}{4}(x - 6)

Multiply both sides by 12: 16x = 36 - 9x + 54, so 25x = 90 and x = \tfrac{18}{5}. Then y = \tfrac{4}{3} \cdot \tfrac{18}{5} = \tfrac{24}{5}. The foot \left(\tfrac{18}{5}, \tfrac{24}{5}\right) is \tfrac{6}{5} times (3, 4), so it lies on the side itself. By the The distance formula,

\begin{aligned} PF &= \sqrt{\left(6 - \tfrac{18}{5}\right)^2 + \left(3 - \tfrac{24}{5}\right)^2} \\ &= \sqrt{\tfrac{144}{25} + \tfrac{81}{25}} = \sqrt{9} = 3 \end{aligned}

Both distances are 3. P is inside the angle: it is above the x-axis, and below the line y = \tfrac{4}{3}x because 3 < \tfrac{4}{3} \cdot 6 = 8. By the converse, \overrightarrow{OP} bisects the angle. It is the ray y = \tfrac{1}{2}x for x \ge 0.

Check. X(5, 0) and Y(3, 4) are both 5 from O, one on each side. Folding along the bisector carries X to Y, so the bisector passes through the midpoint (4, 2) of \overline{XY}. And 2 = \tfrac{1}{2} \cdot 4, so it does.

Example 7

A bisector in a right triangle

In right triangle ABC of Figure 8, the right angle is at C, AC = 12, BC = 16 and AB = 20. The bisector of \angle A meets \overline{BC} at D. Find CD.

Figure 8Example 7. The bisector from A and the perpendicular from D to the hypotenuse.

Solution

Let x = CD, and draw the perpendicular from D to \overline{AB}, with foot E.

D is on the bisector of \angle A, so it is as far from line AB as from line AC. Its distance to line AC is DC = x, because \overline{DC} \perp \overline{AC} (the right angle at C). So DE = x as well.

Triangles ACD and AED are congruent by AAS: \angle CAD \cong \angle EAD (the bisector), the angles at C and E are right, and \overline{AD} is shared. So AE = AC = 12, which leaves EB = 20 - 12 = 8. Also DB = 16 - x.

Triangle DEB has a right angle at E, so by the Pythagorean theorem

x^2 + 8^2 = (16 - x)^2

Expand the right side: x^2 + 64 = 256 - 32x + x^2. The x^2 terms cancel, leaving 32x = 192, so x = 6. That is, CD = 6.

Check. Then DE = 6, EB = 8 and DB = 10, and 6^2 + 8^2 = 100 = 10^2. Notice also that CD : DB = 6 : 10 is the same ratio as AC : AB = 12 : 20. That is no accident. It is a preview of the triangle angle bisector theorem, which proves the ratio for every triangle.

Boundaries

Inside the angle, or the converse fails. Two lines that cross make four angles, and the points equidistant from the two lines fill two whole lines: the bisector of one angle and the bisector of its neighbor. Those two bisectors are perpendicular, because the bisectors of a linear pair always are (Congruent angles, angle addition and bisectors). For the right angle formed by the positive axes, (5, 5) is on its bisector, while (-5, 5) is also 5 from each axis but lies outside the angle, on the line y = -x. That is why the converse asks for a point inside the angle.

Distance means the perpendicular. Equal slanted segments to the two sides prove nothing, as Example 4 showed. Only the perpendicular segment measures the distance.

The angle must be less than a straight angle. If \angle ABC is a straight angle, both sides lie on one line, every point is the same distance from “both” of them, and the converse says nothing. The theorem itself still holds, trivially.

No angle at all. Two parallel lines make no angle and have no bisector. The points equidistant from them form a third parallel line halfway between, which is a fact about the distance between parallel lines.

In space. Out of the plane, the points equidistant from two crossing lines are no longer lines: they fill two planes, each containing one of the bisectors and standing upright on the plane of the two lines.

What a triangle’s bisector does not do. It halves the angle, not the opposite side: in Example 7 it cut a side of 16 into 6 and 10. And the meeting point I is the same distance from the three sides, not from the three vertices. The point the same distance from the vertices comes from the perpendicular bisectors instead.

Where this goes next

This lesson has a twin. The perpendicular bisector theorem says a point is the same distance from two points exactly when it lies on the perpendicular bisector between them; here a point is the same distance from two lines exactly when it lies on an angle bisector. Both are statements about a locus, and both end the same way in a triangle: three special lines meet at one point.

The meeting point of the bisectors is the Incenter. Because it is the same distance from all three sides, a circle centered there with that distance as radius touches each side exactly once, the inscribed circle, and the radius meets each side at a right angle as in Tangent properties. That turns a triangle into a problem about a circle.

Example 7 ended with a ratio. The triangle angle bisector theorem proves that ratio in general: the bisector from a vertex splits the opposite side in the ratio of the two sides that form the angle. And when you construct a bisector with compass and straightedge, the equal-distance property is one way to justify that the ray you drew really is one.

Exercises

Using the theorem

\overrightarrow{BD} bisects \angle ABC, and P is on it. The perpendicular from P to \overrightarrow{BA} has length 11. How long is the perpendicular from P to \overrightarrow{BC}?
P is on the bisector of \angle ABC, with feet F on \overrightarrow{BA} and G on \overrightarrow{BC}. PF = 9 and BF = 12. Find PG, BG and BP.
P is on the bisector of \angle ABC. Its distance to BA is 2x + 3 and its distance to BC is 4x - 7. Find x and the distance.
P is on the bisector of \angle ABC. Its distance to BA is x^2 - 6 and its distance to BC is x. Find x and the distance.
P is on the bisector of \angle ABC, with feet F on \overrightarrow{BA} and G on \overrightarrow{BC}. PG = 15 and BP = 17. Find PF and BF.

Using the converse

P is inside \angle ABC and is the same distance from both sides. m\angle ABP = 37^\circ. Find m\angle PBC and m\angle ABC.
P is inside \angle ABC and is the same distance from both sides. m\angle ABP = (3x + 4)^\circ and m\angle ABC = (5x + 20)^\circ. Find x, m\angle ABP and m\angle ABC.
m\angle ABC = 110^\circ, and P inside the angle is the same distance from both sides. F is the foot of the perpendicular from P to \overrightarrow{BA}. Find m\angle ABP and m\angle BPF.

The bisectors of a triangle

The bisectors of \angle A and \angle B of \triangle ABC meet at I, and I is 6 from \overline{BC}. How far is I from \overline{AB} and from \overline{CA}?
In \triangle ABC, m\angle A = 50^\circ and m\angle B = 70^\circ, and the bisectors of \angle A and \angle B meet at I. Find m\angle IAB, m\angle IBA, m\angle AIB and m\angle ICA.
The angle bisectors of \triangle ABC meet at I, with m\angle IAB = 20^\circ and m\angle IBC = 35^\circ. Find m\angle A, m\angle B, m\angle C and m\angle ICA.
The bisectors of \angle A and \angle B of \triangle ABC meet at I, and m\angle C = 46^\circ. Find m\angle AIB.
The bisectors of \angle A and \angle B of \triangle ABC meet at I, and m\angle AIB = 128^\circ. Find m\angle C.

Bisectors in right triangles

Right triangle ABC has its right angle at C, AC = 6, BC = 8 and AB = 10. The bisector of \angle A meets \overline{BC} at D. Find CD.
In the same triangle as Exercise 14, the bisector of \angle B meets \overline{AC} at K. Find CK.
Right triangle ABC has its right angle at C, AC = 5, BC = 12 and AB = 13. The bisector of \angle A meets \overline{BC} at D. Find CD and AD.

On the coordinate plane

The positive x-axis and the positive y-axis form a right angle at the origin. Find the distance from each of P(7, 7) and Q(3, 8) to both sides. Which point lies on the bisector, and why?
An angle has its vertex at the origin O, one side along the positive x-axis and the other along the ray from O through (5, 12). Show that P(9, 6) is 6 from both sides, and give the slope of the bisector.

Reasoning

P is inside \angle ABC. Its distance to BA is 5 and its distance to BC is 8. Explain why \overrightarrow{BP} is not the bisector of the angle.
R(-5, 5) is 5 from the x-axis and 5 from the y-axis. Is it on the bisector of the right angle formed by the positive axes? Explain which condition of the converse fails, and what R is on instead.
The bisectors of \angle A and \angle B of \triangle ABC meet at I. Prove that m\angle AIB = 90^\circ + \tfrac{1}{2} m\angle C.
Explain why the point where the three bisectors of a triangle meet is always inside the triangle.
Two straight roads cross. Describe every point that is the same distance from both roads, and show that the set is two perpendicular lines.
Explain why the perpendiculars from a point on the bisector of an angle (smaller than a straight angle) always land on the sides themselves, never on their extensions past the vertex.

Answers

11. A point on the bisector is the same distance from both sides.
PG = 9, BG = 12, BP = 15. The theorem gives PG = PF; the congruent triangles PBF and PBG give BG = BF; and \displaystyle BP = \sqrt{9^2 + 12^2} = \sqrt{225} = 15.
x = 5, distance 13. From 2x + 3 = 4x - 7, 10 = 2x. Check: 4(5) - 7 = 13.
x = 3, distance 3. From x^2 - 6 = x, x^2 - x - 6 = 0, so (x - 3)(x + 2) = 0. The root x = -2 would make the distance -2, which no length can be. Check: 3^2 - 6 = 3.
PF = 15, BF = 8. PF = PG by the theorem, and \displaystyle BF = \sqrt{17^2 - 15^2} = \sqrt{64} = 8.
m\angle PBC = 37^\circ, m\angle ABC = 74^\circ. By the converse \overrightarrow{BP} bisects the angle, so the whole is 2 \cdot 37^\circ.
x = 12, m\angle ABP = 40^\circ, m\angle ABC = 80^\circ. By the converse the whole is twice the part: 5x + 20 = 2(3x + 4) = 6x + 8, so x = 12.
m\angle ABP = 55^\circ, m\angle BPF = 35^\circ. By the converse \overrightarrow{BP} halves 110^\circ. 55^\circ is acute, so F is on the side, and triangle BPF has a right angle at F: 180^\circ - 90^\circ - 55^\circ = 35^\circ.
6 from each. On the bisector of \angle B, I is as far from AB as from BC; on the bisector of \angle A, as far from CA as from AB.
25^\circ, 35^\circ, 120^\circ, 30^\circ. The halves are 25^\circ and 35^\circ, and m\angle AIB = 180^\circ - 25^\circ - 35^\circ = 120^\circ. m\angle C = 60^\circ, and \overline{CI} bisects it because all three bisectors meet at I.
m\angle A = 40^\circ, m\angle B = 70^\circ, m\angle C = 70^\circ, m\angle ICA = 35^\circ. Each given angle is half of its vertex’s angle; the angle sum gives 180^\circ - 40^\circ - 70^\circ = 70^\circ, and \overline{CI} halves it.
113^\circ. m\angle A + m\angle B = 180^\circ - 46^\circ = 134^\circ, the halves add to 67^\circ, and m\angle AIB = 180^\circ - 67^\circ = 113^\circ.
76^\circ. The halves at A and B add to 180^\circ - 128^\circ = 52^\circ, so m\angle A + m\angle B = 104^\circ and m\angle C = 180^\circ - 104^\circ = 76^\circ.
CD = 3. With x = CD, the foot E on \overline{AB} has DE = x and AE = AC = 6, so EB = 4 and DB = 8 - x: x^2 + 4^2 = (8 - x)^2 gives 16x = 48.
CK = \tfrac{8}{3}. With y = CK, the foot E on \overline{AB} has KE = y and BE = BC = 8, so AE = 2 and AK = 6 - y: y^2 + 2^2 = (6 - y)^2 gives 12y = 32.
CD = \tfrac{10}{3}, \displaystyle AD = \tfrac{5\sqrt{13}}{3}. With x = CD, AE = 5 and EB = 8: x^2 + 8^2 = (12 - x)^2 gives 24x = 80. Then \displaystyle AD = \sqrt{5^2 + \left(\tfrac{10}{3}\right)^2}, which is \displaystyle \sqrt{\tfrac{325}{9}} = \tfrac{5\sqrt{13}}{3}.
P is 7 from each axis and lies on the bisector y = x: it is inside the angle and equidistant, so the converse applies. Q is 3 from the y-axis and 8 from the x-axis; by the theorem a point of the bisector would be equidistant, so Q is not on it.
The distance to the x-axis is 6. The other side lies on y = \tfrac{12}{5}x; the perpendicular through P has slope -\tfrac{5}{12}, and \tfrac{12}{5}x = 6 - \tfrac{5}{12}(x - 9) gives x = \tfrac{45}{13}, so the foot is \left(\tfrac{45}{13}, \tfrac{108}{13}\right), on the side. The distance is \displaystyle \sqrt{\left(\tfrac{72}{13}\right)^2 + \left(\tfrac{30}{13}\right)^2} = \tfrac{78}{13} = 6. P is inside the angle, so the bisector is \overrightarrow{OP}, with slope \tfrac{6}{9} = \tfrac{2}{3}.
If \overrightarrow{BP} were the bisector, the theorem would make the two distances equal. They are 5 and 8, so it is not.
No. R is equidistant from the two lines but is not inside the angle, since its x-coordinate is negative. It lies on y = -x, the bisector of the neighboring angle formed by the positive y-axis and the negative x-axis.
In \triangle AIB the angles at A and B are \tfrac{1}{2}m\angle A and \tfrac{1}{2}m\angle B, so m\angle AIB = 180^\circ - \tfrac{1}{2}(m\angle A + m\angle B). The angle sum gives m\angle A + m\angle B = 180^\circ - m\angle C, so this is 180^\circ - 90^\circ + \tfrac{1}{2}m\angle C, which is 90^\circ + \tfrac{1}{2}m\angle C.
The bisector from A is a segment from A to a point D of \overline{BC}, so it lies in the triangle. The bisector from B crosses it at a point between A and D, which is inside the triangle, and the third bisector passes through that same point.
The roads make four angles at the crossing. Inside each angle, the equidistant points are its bisector, and the bisectors of opposite angles together form one line. So the set is two lines through the crossing, the bisectors of two neighboring angles. Neighboring angles form a linear pair, and the bisectors of a linear pair meet at \tfrac{1}{2}(180^\circ) = 90^\circ.
Each half of the angle is less than 90^\circ. If the foot were at the vertex, that half would be a right angle; if it were past the vertex, the right triangle formed would also hold the obtuse supplement of that half, and the angles would add to more than 180^\circ.