PAPER · REASONING · PRACTICE

Angle bisectors: perpendicular distances and the forward proof

Distinguish distance from an oblique segment, practise equal lengths, and fade an AAS proof scaffold.

40 minutes · Year 8, upper volume · text pp. 25–26 · foundation lesson 13

Textbook reference: Year 8, upper volume · congruent triangles · angle bisectors

One worked example, one guided attempt, eight core questions, four follow-ups and two extensions. Write on paper; the teacher controls the reveal.

Teacher reference Arrange in My Classroom

Lesson routePhase

4′Check prerequisites
5′Observe and predict
8′Worked example and guided attempt
15′Independent practice
5′Discuss and correct
3′Exit questions

Observe · explain

A question before practising

Keep P on the bisector and compare OP=2,4,6. What stays equal, and what changes? Then move P off the bisector.

Organise the key idea

A point on an internal angle bisector has equal perpendicular distances to the sides. Measurements suggest the result; congruence proves it.

Open the original chapter Open the interactive inquiry

Use the lesson question here for paper discussion; the original activity opens separately.

Worked example1

OP bisects ∠AOB. P is inside the angle, PM⊥OA and PN⊥OB. Prove PM=PN using congruent triangles.

Geometry diagram; not drawn to scale except for construction tasks.

Compare △OPM and △OPN; mark the two right angles.
Find the equal angles at O and the common side OP.
Use the two angle pairs and OP for AAS, then identify the corresponding sides.

∠OMP=∠ONP=90° because the two segments are perpendicular to the sides.
∠MOP=∠PON by the definition of an angle bisector.
OP=OP, the common side. It is opposite the right angle in each triangle.
Therefore △OPM≅△OPN by AAS.
Hence PM=PN, since corresponding sides of congruent triangles are equal.

Check your work

State the right angles and their perpendicular-line reason.
State the equal vertex angles and the common side.
Use a valid congruence criterion without assuming the conclusion.
Keep vertex correspondence consistent and deduce equal distances.
Accept another complete proof if every condition and deduction is justified.

Guided attempt2

OP bisects ∠AOB, P is inside the angle, PM⊥OA and PN⊥OB. Complete the proof table without using PM=PN as an input.

∠OMP=∠___=90° because ___.
∠MOP=∠___ because ___.
___=___ is common, so △OPM≅△OPN by ___.
Therefore ___=___, by ___.

Geometry diagram; not drawn to scale except for construction tasks.

Compare △OPM and △OPN; mark the two right angles.
Find the equal angles at O and the common side OP.
Use the two angle pairs and OP for AAS, then identify the corresponding sides.

∠OMP=∠ONP=90° because the two segments are perpendicular to the sides.
∠MOP=∠PON by the definition of an angle bisector.
OP=OP, the common side. It is opposite the right angle in each triangle.
Therefore △OPM≅△OPN by AAS.
Hence PM=PN, since corresponding sides of congruent triangles are equal.

Check your work

State the right angles and their perpendicular-line reason.
State the equal vertex angles and the common side.
Use a valid congruence criterion without assuming the conclusion.
Keep vertex correspondence consistent and deduce equal distances.
Accept another complete proof if every condition and deduction is justified.

Prerequisite3

M,Q lie on OA. PM⊥OA, while PQ is oblique. Which segment measures the distance from P to OA?

Geometry diagram; not drawn to scale except for construction tasks.

Look for the right-angle mark at the foot.
A segment ending on OA is not automatically perpendicular.
The condition PM⊥OA identifies the required distance.

Distance from a point to a line means the perpendicular length.
PM is perpendicular to OA; OP ends at the vertex and PQ is oblique.

Check your work

Select PM and explain the perpendicular condition.

Prerequisite4

OP bisects ∠AOB=64°. Find ∠AOP.

64°
90°
128°
32°

Geometry diagram; not drawn to scale except for construction tasks.

What does “bisects” mean?
The two equal parts add to 64°.
Divide 64° by 2.

The two parts have equal angles.
64°÷2=32°.

Check your work

Halve the whole angle and retain the degree unit.

Independent practice5

P lies on the internal bisector of ∠AOB; PM⊥OA and PN⊥OB. If PM=6.4 cm, find PN and name the property.

Geometry diagram; not drawn to scale except for construction tasks.

Identify the two perpendicular segments from P.
The angle-bisector property equates their lengths.
Transfer the given length 6.4 to the other perpendicular segment.

Both lengths are perpendicular distances from the same point.
The point lies on the bisector, so PN=PM=6.4 cm.

Check your work

Use perpendicular lengths, not an oblique segment or distance to the vertex.
Give the value with the stated unit and name the property.

Independent practice6

OP is the internal bisector of ∠AOB=74°. Find ∠PON.

Geometry diagram; not drawn to scale except for construction tasks.

The answer is an angle, not a length.
The whole angle is divided into two equal parts.
Calculate 74÷2.

∠AOP=∠PON by the definition of bisector.
Each is 74°÷2=37°.

Check your work

Give 37° with the bisector reason.

Independent practice7

P lies on the internal bisector; PM⊥OA and PN⊥OB. PM=2x+1 cm and PN=3x−2 cm. Find x and check that both lengths are positive.

2x+1=3x-2

Geometry diagram; not drawn to scale except for construction tasks.

First write a geometric equality.
The perpendicular lengths are equal, so equate the expressions.
Solve 2x+1=3x−2, then substitute x into both lengths.

Set PM=PN by the property: 2x+1=3x−2.
Subtract 2x and add 2: x=3.
Substitution gives 2×3+1=3×3−2=7>0.

Check your work

Justify the equation using perpendicular distances.
Solve x=3 and verify both lengths are 7 cm.

Independent practice8

ST bisects ∠CSD and T is inside the angle. TH⊥SC and TK⊥SD. Prove TH=TK; choose the triangles yourself.

Geometry diagram; not drawn to scale except for construction tasks.

Compare △STH and △STK; mark the two right angles.
Find the equal angles at S and the common side ST.
Use the two angle pairs and ST for AAS, then identify the corresponding sides.

∠SHT=∠SKT=90° because the two segments are perpendicular to the sides.
∠HST=∠TSK by the definition of an angle bisector.
ST=ST, the common side. It is opposite the right angle in each triangle.
Therefore △STH≅△STK by AAS.
Hence TH=TK, since corresponding sides of congruent triangles are equal.

Check your work

State the right angles and their perpendicular-line reason.
State the equal vertex angles and the common side.
Use a valid congruence criterion without assuming the conclusion.
Keep vertex correspondence consistent and deduce equal distances.
Accept another complete proof if every condition and deduction is justified.

Exit question9

P lies on the angle bisector and PN⊥OB. Someone gives PQ=3 cm, where Q lies on OA but PQ is not perpendicular to OA. Must PN=3 cm? Explain.

Geometry diagram; not drawn to scale except for construction tasks.

Which condition was changed from the earlier questions?
The theorem uses a perpendicular segment to each side.
PQ is oblique; do not replace PM by PQ.

Check the meaning of the given length before applying the property.
PQ is not a perpendicular distance, so its length cannot determine PN this way.

Check your work

Reject the conclusion and identify the missing perpendicular condition.

Exit question10

A lamp P stands inside the angle formed by two straight paths, on their internal angle bisector. Its shortest distance to the first path is 9 m. Find its shortest distance to the second path.

Geometry diagram; not drawn to scale except for construction tasks.

Identify the two perpendicular segments from P.
The angle-bisector property equates their lengths.
Transfer the given length 9 to the other perpendicular segment.

Interpret each shortest distance as a perpendicular length.
Use the angle-bisector property to transfer 9 m to the second path.

Check your work

Use perpendicular lengths, not an oblique segment or distance to the vertex.
Give the value with the stated unit and name the property.

Further practice11

P lies on the internal bisector, with PM⊥OA and PN⊥OB. PM=3.75 cm. Find PN.

Geometry diagram; not drawn to scale except for construction tasks.

Identify the two perpendicular segments from P.
The angle-bisector property equates their lengths.
Transfer the given length 3.75 to the other perpendicular segment.

PM and PN are the perpendicular distances from P.
Hence PN=PM=3.75 cm.

Check your work

Use perpendicular lengths, not an oblique segment or distance to the vertex.
Give the value with the stated unit and name the property.

Further practice12

P lies on the internal bisector; PM⊥OA and PN⊥OB. PM=3x−1 cm and PN=8 cm. Find x.

Geometry diagram; not drawn to scale except for construction tasks.

Use equal distances first.
Write 3x−1=8.
Add 1, divide by 3, and check the length.

Set 3x−1=8.
Then 3x=9, so x=3.
Check 3×3−1=8>0.

Check your work

Link the equation to the property and obtain x=3.

Further practice13

To prove PM=PN from OP bisecting the angle and the two perpendicular conditions, which starting data are legitimate?

Two right angles, equal angles at O, and OP common.
PM=PN, so HL applies immediately.
Three equal angles prove congruence by AAA.
Assume OM=ON without justification.

Geometry diagram; not drawn to scale except for construction tasks.

Compare △OPM and △OPN; mark the two right angles.
Find the equal angles at O and the common side OP.
Use the two angle pairs and OP for AAS, then identify the corresponding sides.

The perpendicular and bisector conditions supply the two angle pairs.
OP is shared. PM=PN is the target, not a premise.

Check your work

Select only given or independently justified facts.
Explain why assuming PM=PN would be circular.

Further practice14

P is on the internal bisector, PM⊥OA and PN⊥OB. PM=6 cm and △OPM has perimeter 24 cm. Find OP+ON.

Geometry diagram; not drawn to scale except for construction tasks.

The earlier congruence gives another pair of equal sides.
Use OM=ON. Subtract PM from the perimeter.
OP+ON=OP+OM=24−6.

The AAS congruence also gives OM=ON.
OP+OM=24−6=18 cm.
Replace OM by ON to obtain OP+ON=18 cm.

Check your work

Justify OM=ON and subtract the correct side.

Extension15

Three trials show the same rounded values for PM and PN. Explain what this supports, and what is still needed for every point on the bisector.

Geometry diagram; not drawn to scale except for construction tasks.

How many points did the trials test?
A screen displays rounded measurements.
Connect the general conditions to AAS.

A finite set of trials covers only those settings.
Rounded equality may conceal a small difference.
The AAS proof uses the stated conditions for an arbitrary eligible point.

Check your work

Distinguish conjectural evidence from a general proof.

Extension16

A pupil says every point inside an angle has equal distances to the sides. A point Q inside the angle has perpendicular distances 2 cm and 4 cm. Is it a counterexample, and what condition repairs the claim?

A counterexample must satisfy the premise and violate the conclusion.
Compare 2 and 4.
“Inside the angle” alone is weaker than “on the angle bisector”.

Q satisfies the proposed premise, being inside the angle.
Its perpendicular distances differ, so the proposed conclusion fails.
The forward theorem requires the point to be on the bisector.

Check your work

Verify both parts of the counterexample and state the missing bisector condition.

Preparation and assessment notes

Prerequisites

Angle-bisector definition, perpendicular segments, AAS, and simple linear equations. P is an interior point different from O; M and N are the perpendicular feet on the sides.

Look for in students’ work

Use a short projected observation, one complete AAS model, and a fill-in attempt before independent work. Never introduce PM=PN as a given in the forward proof. Accept other justified proofs.

Collect the paper exit questions and note completion and common errors. This page does not grade or store paper answers.

Paper for the class

Download all worksheets (ZIP · 318 PDFs · 47.2 MB)

All 53 Year 7–8 upper- and lower-volume packs in Traditional Chinese and English. Includes teacher answers; student core, further practice and teacher papers are in separate folders.

Core student worksheet: two A4 pages. Further practice is separate. Teacher copies include reference reasoning.

Print edition

Textbook reference and authorship

Original practice aligned to the referenced skills. Use your own textbook for the cited pages.

大湾区初中数学三维课堂（八年级上册）正文.pdf

Year 8, upper volume · text pp. 25–26 · foundation lesson 13

Original questions and explanations; the source textbook and answer PDFs are not reproduced here.

Angle bisectors: perpendicular distances and the forward proof

Year 8, upper volume · text pp. 25–26 · foundation lesson 13 · 1/2

Name: __________ Class: ______ Date: __________

Worked example1

OP bisects ∠AOB. P is inside the angle, PM⊥OA and PN⊥OB. Prove PM=PN using congruent triangles.

Geometry diagram; not drawn to scale except for construction tasks.

Guided attempt2

OP bisects ∠AOB, P is inside the angle, PM⊥OA and PN⊥OB. Complete the proof table without using PM=PN as an input.

∠OMP=∠___=90° because ___.
∠MOP=∠___ because ___.
___=___ is common, so △OPM≅△OPN by ___.
Therefore ___=___, by ___.

Geometry diagram; not drawn to scale except for construction tasks.

Prerequisite3

M,Q lie on OA. PM⊥OA, while PQ is oblique. Which segment measures the distance from P to OA?

Geometry diagram; not drawn to scale except for construction tasks.

Prerequisite4

OP bisects ∠AOB=64°. Find ∠AOP.

64°
90°
128°
32°

Geometry diagram; not drawn to scale except for construction tasks.

Angle bisectors: perpendicular distances and the forward proof

Year 8, upper volume · text pp. 25–26 · foundation lesson 13 · 2/2

Name: __________ Class: ______ Date: __________

Independent practice5

P lies on the internal bisector of ∠AOB; PM⊥OA and PN⊥OB. If PM=6.4 cm, find PN and name the property.

Geometry diagram; not drawn to scale except for construction tasks.

Independent practice6

OP is the internal bisector of ∠AOB=74°. Find ∠PON.

Geometry diagram; not drawn to scale except for construction tasks.

Independent practice7

P lies on the internal bisector; PM⊥OA and PN⊥OB. PM=2x+1 cm and PN=3x−2 cm. Find x and check that both lengths are positive.

2x+1=3x-2

Geometry diagram; not drawn to scale except for construction tasks.

Independent practice8

ST bisects ∠CSD and T is inside the angle. TH⊥SC and TK⊥SD. Prove TH=TK; choose the triangles yourself.

Geometry diagram; not drawn to scale except for construction tasks.

Exit question9

P lies on the angle bisector and PN⊥OB. Someone gives PQ=3 cm, where Q lies on OA but PQ is not perpendicular to OA. Must PN=3 cm? Explain.

Geometry diagram; not drawn to scale except for construction tasks.

Exit question10

A lamp P stands inside the angle formed by two straight paths, on their internal angle bisector. Its shortest distance to the first path is 9 m. Find its shortest distance to the second path.

Geometry diagram; not drawn to scale except for construction tasks.

Angle bisectors: perpendicular distances and the forward proof — Further practice

Year 8, upper volume · text pp. 25–26 · foundation lesson 13

Further practice11

P lies on the internal bisector, with PM⊥OA and PN⊥OB. PM=3.75 cm. Find PN.

Geometry diagram; not drawn to scale except for construction tasks.

Further practice12

P lies on the internal bisector; PM⊥OA and PN⊥OB. PM=3x−1 cm and PN=8 cm. Find x.

Geometry diagram; not drawn to scale except for construction tasks.

Further practice13

To prove PM=PN from OP bisecting the angle and the two perpendicular conditions, which starting data are legitimate?

Two right angles, equal angles at O, and OP common.
PM=PN, so HL applies immediately.
Three equal angles prove congruence by AAA.
Assume OM=ON without justification.

Geometry diagram; not drawn to scale except for construction tasks.

Further practice14

P is on the internal bisector, PM⊥OA and PN⊥OB. PM=6 cm and △OPM has perimeter 24 cm. Find OP+ON.

Geometry diagram; not drawn to scale except for construction tasks.

Extension15

Three trials show the same rounded values for PM and PN. Explain what this supports, and what is still needed for every point on the bisector.

Geometry diagram; not drawn to scale except for construction tasks.

Extension16

Angle bisectors: perpendicular distances and the forward proof — Teacher answers

Year 8, upper volume · text pp. 25–26 · foundation lesson 13

Use a short projected observation, one complete AAS model, and a fill-in attempt before independent work. Never introduce PM=PN as a given in the forward proof. Accept other justified proofs.

Worked example1

OP bisects ∠AOB. P is inside the angle, PM⊥OA and PN⊥OB. Prove PM=PN using congruent triangles.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

△OPM≅△OPN by AAS, so PM=PN.

∠OMP=∠ONP=90° because the two segments are perpendicular to the sides.
∠MOP=∠PON by the definition of an angle bisector.
OP=OP, the common side. It is opposite the right angle in each triangle.
Therefore △OPM≅△OPN by AAS.
Hence PM=PN, since corresponding sides of congruent triangles are equal.

State the right angles and their perpendicular-line reason.
State the equal vertex angles and the common side.
Use a valid congruence criterion without assuming the conclusion.
Keep vertex correspondence consistent and deduce equal distances.
Accept another complete proof if every condition and deduction is justified.

Guided attempt2

OP bisects ∠AOB, P is inside the angle, PM⊥OA and PN⊥OB. Complete the proof table without using PM=PN as an input.

∠OMP=∠___=90° because ___.
∠MOP=∠___ because ___.
___=___ is common, so △OPM≅△OPN by ___.
Therefore ___=___, by ___.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

Right angles OMP and ONP; ∠MOP=∠PON; OP=OP; AAS; PM=PN.

∠OMP=∠ONP=90° because the two segments are perpendicular to the sides.
∠MOP=∠PON by the definition of an angle bisector.
OP=OP, the common side. It is opposite the right angle in each triangle.
Therefore △OPM≅△OPN by AAS.
Hence PM=PN, since corresponding sides of congruent triangles are equal.

State the right angles and their perpendicular-line reason.
State the equal vertex angles and the common side.
Use a valid congruence criterion without assuming the conclusion.
Keep vertex correspondence consistent and deduce equal distances.
Accept another complete proof if every condition and deduction is justified.

Prerequisite3

M,Q lie on OA. PM⊥OA, while PQ is oblique. Which segment measures the distance from P to OA?

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

PM: the perpendicular segment from the point to the side.

Distance from a point to a line means the perpendicular length.
PM is perpendicular to OA; OP ends at the vertex and PQ is oblique.

Select PM and explain the perpendicular condition.

Prerequisite4

OP bisects ∠AOB=64°. Find ∠AOP.

64°
90°
128°
32°

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

32°.

The two parts have equal angles.
64°÷2=32°.

Halve the whole angle and retain the degree unit.

Independent practice5

P lies on the internal bisector of ∠AOB; PM⊥OA and PN⊥OB. If PM=6.4 cm, find PN and name the property.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

PN=6.4 cm, by the angle-bisector property.

Both lengths are perpendicular distances from the same point.
The point lies on the bisector, so PN=PM=6.4 cm.

Use perpendicular lengths, not an oblique segment or distance to the vertex.
Give the value with the stated unit and name the property.

Independent practice6

OP is the internal bisector of ∠AOB=74°. Find ∠PON.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

∠PON=37°.

∠AOP=∠PON by the definition of bisector.
Each is 74°÷2=37°.

Give 37° with the bisector reason.

Independent practice7

P lies on the internal bisector; PM⊥OA and PN⊥OB. PM=2x+1 cm and PN=3x−2 cm. Find x and check that both lengths are positive.

2x+1=3x-2

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

x=3; both lengths are 7 cm.

Set PM=PN by the property: 2x+1=3x−2.
Subtract 2x and add 2: x=3.
Substitution gives 2×3+1=3×3−2=7>0.

Justify the equation using perpendicular distances.
Solve x=3 and verify both lengths are 7 cm.

Independent practice8

ST bisects ∠CSD and T is inside the angle. TH⊥SC and TK⊥SD. Prove TH=TK; choose the triangles yourself.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

△STH≅△STK by AAS; hence TH=TK.

∠SHT=∠SKT=90° because the two segments are perpendicular to the sides.
∠HST=∠TSK by the definition of an angle bisector.
ST=ST, the common side. It is opposite the right angle in each triangle.
Therefore △STH≅△STK by AAS.
Hence TH=TK, since corresponding sides of congruent triangles are equal.

State the right angles and their perpendicular-line reason.
State the equal vertex angles and the common side.
Use a valid congruence criterion without assuming the conclusion.
Keep vertex correspondence consistent and deduce equal distances.
Accept another complete proof if every condition and deduction is justified.

Exit question9

P lies on the angle bisector and PN⊥OB. Someone gives PQ=3 cm, where Q lies on OA but PQ is not perpendicular to OA. Must PN=3 cm? Explain.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

No. PQ is oblique and does not measure the distance to OA. The property equates perpendicular distances only.

Check the meaning of the given length before applying the property.
PQ is not a perpendicular distance, so its length cannot determine PN this way.

Reject the conclusion and identify the missing perpendicular condition.

Exit question10

A lamp P stands inside the angle formed by two straight paths, on their internal angle bisector. Its shortest distance to the first path is 9 m. Find its shortest distance to the second path.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

9 m: shortest distances are perpendicular distances, and the point is on the bisector.

Interpret each shortest distance as a perpendicular length.
Use the angle-bisector property to transfer 9 m to the second path.

Use perpendicular lengths, not an oblique segment or distance to the vertex.
Give the value with the stated unit and name the property.

Further practice11

P lies on the internal bisector, with PM⊥OA and PN⊥OB. PM=3.75 cm. Find PN.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

PN=3.75 cm.

PM and PN are the perpendicular distances from P.
Hence PN=PM=3.75 cm.

Use perpendicular lengths, not an oblique segment or distance to the vertex.
Give the value with the stated unit and name the property.

Further practice12

P lies on the internal bisector; PM⊥OA and PN⊥OB. PM=3x−1 cm and PN=8 cm. Find x.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

x=3; both perpendicular lengths are 8 cm.

Set 3x−1=8.
Then 3x=9, so x=3.
Check 3×3−1=8>0.

Link the equation to the property and obtain x=3.

Further practice13

To prove PM=PN from OP bisecting the angle and the two perpendicular conditions, which starting data are legitimate?

Two right angles, equal angles at O, and OP common.
PM=PN, so HL applies immediately.
Three equal angles prove congruence by AAA.
Assume OM=ON without justification.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

Use the right angles, the bisected angles, and common OP; AAS proves the result.

The perpendicular and bisector conditions supply the two angle pairs.
OP is shared. PM=PN is the target, not a premise.

Select only given or independently justified facts.
Explain why assuming PM=PN would be circular.

Further practice14

P is on the internal bisector, PM⊥OA and PN⊥OB. PM=6 cm and △OPM has perimeter 24 cm. Find OP+ON.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

OP+ON=18 cm.

The AAS congruence also gives OM=ON.
OP+OM=24−6=18 cm.
Replace OM by ON to obtain OP+ON=18 cm.

Justify OM=ON and subtract the correct side.

Extension15

Three trials show the same rounded values for PM and PN. Explain what this supports, and what is still needed for every point on the bisector.

Geometry diagram; not drawn to scale except for construction tasks.

Reference answer

The trials support the conjecture. A general congruence proof, not repeated or rounded measurements, is needed for all such points.

A finite set of trials covers only those settings.
Rounded equality may conceal a small difference.
The AAS proof uses the stated conditions for an arbitrary eligible point.

Distinguish conjectural evidence from a general proof.

Extension16

Reference answer

Yes: 2≠4. Restrict the point to the internal angle bisector.

Q satisfies the proposed premise, being inside the angle.
Its perpendicular distances differ, so the proposed conclusion fails.
The forward theorem requires the point to be on the bisector.

Verify both parts of the counterexample and state the missing bisector condition.