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Angle Bisectors and the Incenter: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Core (core-course level)

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Problem 1 of 10

Problem 1 Lengths from an angle-bisector point

P lies on the internal bisector of angle UOV. X and Y are the perpendicular feet on its side lines. The given lengths in centimeters are
$\begin{aligned}PX&=3x+2\\PY&=5x-10\\OP&=29.\end{aligned}$
Find x, both perpendicular distances, and OX. Check that the lengths can form the indicated right triangle.

P is on the internal bisector, with perpendicular feet X and Y.

Text description of this figure

P lies on the internal bisector of angle UOV. X and Y are perpendicular feet on the forward side rays. PX is 3x plus 2 centimeters, PY is 5x minus 10 centimeters, and OP is 29 centimeters.
1. Hint 1
  First connect the two perpendicular lengths using the angle-bisector theorem.
2. Hint 2
  After finding the common perpendicular distance, identify the hypotenuse in triangle OXP.
Answer

$\begin{aligned}x&=6\\PX&=20\text{ cm}\\PY&=20\text{ cm}\\OX&=21\text{ cm}.\end{aligned}$
The positive legs and hypotenuse form a right triangle.

Full solution

The perpendicular lengths are equal.
$\begin{aligned}3x+2&=5x-10\\12&=2x\\x&=6.\end{aligned}$

Substitute into the two lengths separately.
$\begin{aligned}PX&=3(6)+2\\PX&=20\text{ cm}\\PY&=5(6)-10\\PY&=20\text{ cm}.\end{aligned}$

OP is the hypotenuse in right triangle OXP.
$\begin{aligned}OX^2&=29^2-20^2\\OX^2&=441\\OX&=21\text{ cm}.\end{aligned}$

The lengths are positive, and the check is
$20^2+21^2=29^2.$
Two reflected copies about OP make a valid angle with P on its internal bisector.

Answer

$\begin{aligned}x&=6\\PX&=20\text{ cm}\\PY&=20\text{ cm}\\OX&=21\text{ cm}.\end{aligned}$
The positive legs and hypotenuse form a right triangle.

Key idea

Use equal perpendiculars first, then the right triangle they create.
Problem 2 Read three angle halves

I is the incenter of triangle ABC. The given angles are
$\begin{aligned}\angle A&=48^\circ\\\angle B&=76^\circ.\end{aligned}$
Find $\angle ACI$ and $\angle BIC$ . Explain why halving the angle at C alone is not enough to find the angle at I.

The three pairs of angle halves use different arc patterns.

Text description of this figure

Angle A is 48 degrees and angle B is 76 degrees. I is the given incenter. Matching patterns are confined to the two halves at each vertex.
1. Hint 1
  Use the angle sum of the original triangle before considering a smaller triangle.
2. Hint 2
  In triangle BIC, the angles at B and C are halves of the original vertex angles.
Answer

$\begin{aligned}\angle ACI&=28^\circ\\\angle BIC&=114^\circ.\end{aligned}$
The angle at I requires the angle sum in triangle BIC.

Full solution

The remaining vertex angle is
$\begin{aligned}\angle C&=180^\circ-48^\circ-76^\circ\\\angle C&=56^\circ.\end{aligned}$

CI bisects the angle at C.
$\angle ACI=28^\circ.$

The angles at B and C in triangle BIC are 38 degrees and 28 degrees.
$\begin{aligned}\angle BIC&=180^\circ-38^\circ-28^\circ\\\angle BIC&=114^\circ.\end{aligned}$

Halving the angle at C gives an angle at C.
The angle at I belongs to the smaller triangle and must be found from its own three-angle sum.

Answer

$\begin{aligned}\angle ACI&=28^\circ\\\angle BIC&=114^\circ.\end{aligned}$
The angle at I requires the angle sum in triangle BIC.

Key idea

Keep track of which triangle contains the angle you are finding.
Problem 3 Prove the forward distance theorem

P is a point other than O on the internal bisector of angle UOV, whose measure is strictly between zero and 180 degrees. Perpendiculars from P meet the side rays at X and Y.

Prove PX equals PY using triangle congruence. Identify the angle pairs, the shared side, and the congruence test. Explain why this proof does not need OU to equal OV.

P lies on the internal bisector; X and Y are perpendicular feet.

Text description of this figure

P lies on the internal bisector of angle UOV. PX and PY are perpendicular to its two side lines. Only the given half-angle equality is marked.
1. Hint 1
  Compare triangles OXP and OYP.
2. Hint 2
  The shared segment is a hypotenuse, and the internal bisector supplies an angle pair.
Answer

PX equals PY by AAS; the proof uses OP, not the arbitrary ray lengths OU and OV.

Full solution

The perpendiculars give right angles at X and Y.
The internal bisector gives equal angles XOP and POY.

The side OP belongs to both triangles.
These two angle pairs and this nonincluded side give congruence by AAS.

Corresponding legs PX and PY are equal.
No step used U or V as a triangle vertex, so their distances from O need not agree.

Answer

PX equals PY by AAS; the proof uses OP, not the arbitrary ray lengths OU and OV.

Key idea

Match only the objects actually used in the congruent triangles.
Problem 4 Build and annotate a bisector

Draw an obtuse angle of your choice, strictly less than 180 degrees. Construct its internal angle bisector using only a compass and straightedge. Leave the arcs visible and label the vertex O, the first-arc points U,V, and the interior crossing X.

Annotate the equal lengths that justify the construction, and write a short congruence proof. Explain which crossing to use if the later arcs also pass through O.
1. Hint 1
  The first arc creates one pair of equal radii.
2. Hint 2
  Equal later settings create another pair. Include the side shared by the two triangles.
Answer

A valid construction uses the non-O crossing X inside the angle and ray OX. The SSS pairs are
$\begin{aligned}OU&=OV\\UX&=VX\\OX&=OX.\end{aligned}$

Full solution

Draw an arc centered at O meeting the rays at distinct points U,V.
Keep its radius and draw arcs centered at U and V.
They pass through O and meet again at an interior point X.

Use the second, interior crossing X, since choosing O would not determine a new ray.
Draw ray OX.

Mark OU and OV equal as first-circle radii.
Mark UX and VX equal as the later equal radii.
When the same setting is kept throughout, all four marked lengths are equal.

Triangles OUX and OVX have two equal side pairs and shared side OX.
SSS makes them congruent, so their corresponding angles at O are equal.
X is inside the chosen angle, so OX is its internal bisector.

One valid obtuse-angle construction, with its SSS lengths marked.

Answer

A valid construction uses the non-O crossing X inside the angle and ray OX. The SSS pairs are
$\begin{aligned}OU&=OV\\UX&=VX\\OX&=OX.\end{aligned}$

Key idea

A construction is justified by equalities created by the tool settings.
Problem 5 Reverse the theorem with the right hypotheses

P lies strictly inside angle UOV, whose measure is below 180 degrees. The perpendicular feet X,Y lie on the forward side rays. PX equals PY.

Prove that ray OP is the internal bisector. Then explain why the same congruence argument would need a position check if a foot were only known to be on a full supporting line.

Equal perpendiculars are given; equal angle halves are to be proved.

Text description of this figure

P is inside angle UOV. PX and PY are perpendicular to the side lines and have equal lengths. No equal-angle marks are supplied.
1. Hint 1
  Name the common hypotenuse and the given matching legs.
2. Hint 2
  After proving congruence, check whether the matching angles are parts of the specified angle or involve backward rays.
Answer

HL gives equal interior angles XOP and POY, so OP is the internal bisector; full-line feet require a forward-ray position check.

Full solution

Triangles OXP and OYP are right triangles.
OP is their common hypotenuse, and the corresponding legs PX and PY are equal.
HL proves the triangles congruent.

Corresponding vertex angles XOP and POY are therefore equal.
Because X,Y are on the forward rays and P is inside, these are exactly the two parts of angle UOV.

A foot on a backward extension would make an angle of the right triangle use that backward ray.
Equal acute angles in those right triangles cannot be identified with the two interior angle parts without checking positions.

Answer

HL gives equal interior angles XOP and POY, so OP is the internal bisector; full-line feet require a forward-ray position check.

Key idea

Congruence gives equal angles; incidence identifies which angles they are.
Problem 6 Construct the incircle on paper

Draw a nondegenerate scalene triangle ABC. Construct its incircle with compass and straightedge. Leave enough construction marks to show how you found the center I and the radius.

Label the perpendicular feet D,E,F on BC,CA,AB. Explain why they lie on the side segments, why their distances from I agree, and why a radius to a vertex is not the required compass setting.
1. Hint 1
  Locate the center by intersecting two internal angle bisectors.
2. Hint 2
  Construct a perpendicular from the center to one side and use its length as radius.
3. Hint 3
  Use the two endpoint angles of a side to locate its foot within the segment.
Answer

The incircle has center I, the intersection of two internal bisectors, and radius IF; its three perpendicular feet lie on the sides and have equal distances from I.

Full solution

Construct the internal bisectors at A and B and name their interior intersection I.
Construct IF perpendicular to AB and set the compass to IF.
Draw the circle centered at I.

Construct the other perpendicular feet D and E for annotation.
The bisector theorems at A and B give, respectively,
$\begin{aligned}IF&=IE\\IF&=ID.\end{aligned}$
Thus all three perpendicular lengths are the same radius.

For AB, the acute half-angle at A puts F on ray AB, and the acute half-angle at B puts F on ray BA.
Hence F lies strictly between A and B.
Apply the same argument to the other two sides, using the third bisector supplied by concurrency.

Each side line is at distance IF from I, so the circle reaches it only at its perpendicular foot.
Any other point of that line is farther from I by Pythagoras.
The circle therefore touches the three side segments.

A segment from I to a vertex is a hypotenuse beside a perpendicular radius, so it is longer than the inradius.
It would give the wrong compass setting.

A completed construction, with the radius to each side.

Answer

The incircle has center I, the intersection of two internal bisectors, and radius IF; its three perpendicular feet lie on the sides and have equal distances from I.

Key idea

Two bisectors locate the center; a perpendicular supplies the radius.
Problem 7 Compare an internal ray with an external bisector

Angle UOV measures 120 degrees and opens to the right as shown. P lies inside it and Q lies above O outside it. Each point is equidistant from the two supporting side lines. The one-tick lengths belong to P; the two-tick lengths belong to Q.

Which of OP and OQ is the internal angle-bisector ray? Explain why the equal-distance fact has a different conclusion for Q, and identify the foot on a backward extension.

The dashed backward extension contains one of Q’s feet.

Text description of this figure

P lies inside the 120-degree angle UOV and Q lies outside above O. Each has equal perpendicular distances to the two supporting lines. The two points have distinct tick patterns; their distances are not asserted equal to each other.
1. Hint 1
  Check the interior-point hypothesis before applying the converse.
2. Hint 2
  Trace each perpendicular foot back to its original forward ray.
Answer

OP is the internal bisector; OQ follows an external bisector line, and QW ends on the backward extension of OV.

Full solution

P is inside the angle and its two perpendicular distances are equal.
The interior-point converse places OP on the internal bisector.

Q is outside the chosen angle, so that converse cannot identify OQ as its internal ray.
The two supporting lines also have an external bisector, which contains Q here.

Z is on the forward ray OU, but W is on line OV behind O.
Thus QW measures distance to that full line, not distance to the forward side ray.

Answer

OP is the internal bisector; OQ follows an external bisector line, and QW ends on the backward extension of OV.

Key idea

The chosen angle and its interior select a ray from the full-line equal-distance possibilities.
Problem 8 Repair an unrestricted equal-distance claim

The displayed right triangle ABC has AB of length 4 and AC of length 3. Q is outside it. D,E,F are perpendicular feet on the full lines BC,CA,AB, respectively, and all three distances from Q are 6.

Consider the claim, "Any point equally far from all three supporting side lines is the incenter." Decide whether the claim is valid. Use the given configuration to explain your decision and repair the claim with a sufficient location condition.

The hypothesis explicitly permits full-line feet beyond vertices.

Text description of this figure

Q is outside right triangle ABC. AB is 4 and AC is 3. QF,QE,QD are perpendicular to lines AB,AC,BC and each has length 6. F and E lie on extensions; D lies on BC.
1. Hint 1
  An incenter has an interior-location requirement as well as an equal-distance property.
2. Hint 2
  Decide whether the converse can select internal bisectors at every vertex for this exterior Q.
Answer

The claim is false; Q is an exterior counterexample. Requiring the point to lie strictly inside the triangle makes equal distances to the three side lines sufficient.

Full solution

The three supplied perpendicular distances agree, but Q is outside the triangle.
The incenter lies strictly inside, so Q is not the incenter.
This is a counterexample to the unrestricted claim.

Here F lies beyond B and E lies beyond C.
The full-line distance condition includes these extensions.
It therefore allows an exterior center as well as the triangle’s interior incenter.

Repair the claim by requiring the point to be strictly inside the nondegenerate triangle.
It then lies inside each vertex angle, so the interior-point converse applies at A and B.
The point is on both internal bisectors and hence is their unique intersection, the incenter.

Answer

The claim is false; Q is an exterior counterexample. Requiring the point to lie strictly inside the triangle makes equal distances to the three side lines sufficient.

Key idea

Equal distances to supporting lines require an interior condition to select the incenter.
Problem 9 Show that the conditions identify only one point

P and Q are strictly inside the same nondegenerate triangle ABC. Each is equidistant from the three supporting side lines, but their common distances are not initially assumed equal to each other.

Show that P and Q must be the same point. Explain why knowing equal distances to AB and AC alone would not give the same uniqueness conclusion.
1. Hint 1
  Apply the interior-point converse separately at two vertices to each candidate.
2. Hint 2
  How many common points can two distinct internal-bisector lines have?
Answer

P and Q coincide at the incenter; one equal-distance pair alone leaves an entire internal-bisector ray of candidates inside the angle.

Full solution

For P, equal distances to AB and AC put it on the internal bisector at A.
Equal distances to AB and BC put it on the internal bisector at B.
Its interior location supplies both converse hypotheses.

The same argument puts Q on those same two internal bisectors.
These lines are distinct: line AB itself cannot bisect either nonzero vertex angle internally.

Two distinct lines have at most one intersection, so P and Q are the same point.
Their common distance values therefore agree as a consequence, without needing to assume that equality.

Equal distances to AB and AC alone put a point on the internal bisector at A.
Many points of that ray lie inside the triangle, so that single pair does not locate a unique center.

Answer

P and Q coincide at the incenter; one equal-distance pair alone leaves an entire internal-bisector ray of candidates inside the angle.

Key idea

Two independent interior-bisector conditions supply uniqueness.
Problem 10 Transfer the distance test to a rectangle

ABCD is a rectangle with AB of length 12 and AD of length 8. Can one circle lying inside it touch all four sides? Justify your answer using perpendicular distances from a possible center to opposite sides.

Explain why the triangle incenter theorem does not by itself answer the question for a rectangle. A drawing of a circle that almost fits is not a justification.

Rectangle ABCD has adjacent side lengths 12 and 8.

Text description of this figure

ABCD is a rectangle with AB of length 12 and AD of length 8. P is an arbitrary interior point. Perpendiculars from P to the four sides are drawn without equality marks.
1. Hint 1
  If a circle touches a side, its center-to-side perpendicular distance must equal its radius.
2. Hint 2
  The two perpendicular distances to each pair of opposite sides add to the distance between those sides.
Answer

No such circle exists. One pair of opposite sides would require radius 4 and the other radius 6; the triangle theorem only guarantees the construction for triangles.

Full solution

A circle contained in the rectangle cannot have a radius larger than a center-to-side perpendicular distance: the radius pointing directly toward that side would cross it.
To reach the side, its radius cannot be smaller than that shortest distance either.
Thus touching while staying inside requires equality.

Suppose an interior circle touches all four sides and has radius r.
The perpendicular distances from its center to AD and BC are both r.
These opposite sides are 12 units apart.

The horizontal perpendicular lengths add across the rectangle.
$\begin{aligned}2r&=12\\r&=6.\end{aligned}$

The distances to AB and CD must also both be r, but those sides are 8 units apart.
$\begin{aligned}2r&=8\\r&=4.\end{aligned}$

One circle cannot have both radii, so the required circle does not exist.
The triangle theorem proves concurrency and equal distances for three triangle sides; it gives no guarantee for four rectangle sides.

Answer

No such circle exists. One pair of opposite sides would require radius 4 and the other radius 6; the triangle theorem only guarantees the construction for triangles.

Key idea

A method transferred to a new shape must satisfy every new side condition.

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