PAPER · REASONING · PRACTICE

Perpendicular bisectors: properties and the SSS converse — Teacher reference

Separate perpendicularity from midpoint information, prove equal endpoint distances, and use SSS for the converse.

40 minutes · Year 8, upper volume · text pp. 31–32 · foundation lesson 16

Textbook reference: Year 8, upper volume · reflection symmetry

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

Student classroom page 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

Compare two candidate lines: one perpendicular but missing the midpoint, and one through the midpoint but slanted. Which definition requirement does each fail?

Organise the key idea

The perpendicular bisector passes through a segment’s midpoint and is perpendicular to it. Its points are equidistant from the endpoints, and the converse also holds.

Open the original chapter

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

Worked example1

M is the midpoint of AB. P is off AB and PM⊥AB. Prove PA=PB.

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

Compare △PMA and △PMB.
Use the midpoint for equal half-segments and perpendicularity for the included right angles.
The second side is common PM; SAS gives PA=PB.

AM=BM because M is the midpoint of AB.
∠PMA=∠PMB=90° because PM⊥AB.
PM=PM is common, so △PMA≅△PMB by SAS.
Thus PA=PB, by corresponding sides of congruent triangles.

Check your work

Use the midpoint and perpendicular conditions separately.
State the common side and justify the congruence criterion.
Deduce the target without assuming equal endpoint distances first.
Accept another complete proof if every condition and deduction is justified.

Guided attempt2

H is the midpoint of CD, Q is off CD and QH⊥CD. Complete a proof that QC=QD.

CH=___ because H is ___.
∠QHC=∠___=___ because ___.
___ is common; the criterion is ___.
Therefore QC=QD by ___.

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

Compare △QHC and △QHD.
Use the midpoint for equal half-segments and perpendicularity for the included right angles.
The second side is common QH; SAS gives QC=QD.

CH=DH because H is the midpoint of CD.
∠QHC=∠QHD=90° because QH⊥CD.
QH=QH is common, so △QHC≅△QHD by SAS.
Thus QC=QD, by corresponding sides of congruent triangles.

Check your work

Use the midpoint and perpendicular conditions separately.
State the common side and justify the congruence criterion.
Deduce the target without assuming equal endpoint distances first.
Accept another complete proof if every condition and deduction is justified.

Prerequisite3

M is the midpoint of AB and AB=10 cm. Find AM.

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

Recall the definition of midpoint.
The whole length consists of two equal halves.
Divide 10 by 2.

The midpoint divides AB into two equal segments.
10÷2=5, so AM=5 cm.

Check your work

Use the correct midpoint or perpendicular-bisector property and retain cm.

Prerequisite4

A,M,B are collinear, P is off AB and PM⊥AB. What is ∠PMA?

45°
180°
It must equal the length AM.
90°

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

Locate the angle’s vertex M.
One ray lies on AB and the other lies on PM.
Perpendicular lines form right angles of 90°.

MA is part of the line AB.
PM⊥AB therefore makes ∠PMA a right angle.

Check your work

Give an angle of 90°, not a length.

Independent practice5

M is the midpoint of AB and AB=18 cm. Find AM.

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

Recall the definition of midpoint.
The whole length consists of two equal halves.
Divide 18 by 2.

The midpoint divides AB into two equal segments.
18÷2=9, so AM=9 cm.

Check your work

Use the correct midpoint or perpendicular-bisector property and retain cm.

Independent practice6

P lies on the perpendicular bisector of AB. If PA=8 cm, find PB and state the property.

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

Identify the segment whose perpendicular bisector contains P.
Compare distances to the two endpoints, not to the line itself.
Use PB=PA=8.

The point is on the perpendicular bisector, so PA=PB.
Substitute PA=8 cm to obtain PB=8 cm.

Check your work

Use the correct midpoint or perpendicular-bisector property and retain cm.

Independent practice7

P lies on the perpendicular bisector of AB. PA=3x+1 cm and PB=2x+5 cm. Find x and check both distances.

3x+1=2x+5

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

Which endpoint distances does this property equate?
Set PA=PB before solving.
Solve 3x+1=2x+5 and substitute into both expressions.

The perpendicular-bisector property gives PA=PB, so 3x+1=2x+5.
Subtract 2x and then subtract 1: x=4.
Check: 3×4+1=13 and 2×4+5=13; both are positive.

Check your work

Justify equal endpoint distances, solve x=4, and verify 13 cm.

Independent practice8

P is off AB and PA=PB. M is the midpoint of AB. Prove PM⊥AB, and conclude that P lies on the perpendicular bisector of AB.

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

Compare triangles PMA and PMB using three side pairs.
SSS gives equal angles at M, but equality alone does not yet mean 90°.
Use the straight line A–M–B: the two equal angles sum to 180°.

PA=PB is given, AM=BM because M is the midpoint, and PM is common.
Therefore △PMA≅△PMB by SSS.
Corresponding angles give ∠PMA=∠PMB.
A,M,B are collinear, so these adjacent angles total 180°; each is 90°.
Thus PM⊥AB through its midpoint, placing P on the perpendicular bisector.

Check your work

Supply all three SSS side pairs.
Use both equal angles and their straight-angle sum.
Conclude perpendicularity through the midpoint.
Accept another complete proof if every condition and deduction is justified.

Exit question9

A line passes through the midpoint of AB, but it is not stated to be perpendicular to AB. Must every point on it be equidistant from A and B?

No; the perpendicular condition is also needed.
Yes; midpoint information alone is enough.
Yes; all lines through one point coincide.
No line can be perpendicular to AB.

Split the term “perpendicular bisector” into two requirements.
One requirement is passing through the midpoint.
The other is being perpendicular to AB.

Passing through the midpoint supplies the “bisector” part only.
Without perpendicularity, the equal-distance theorem does not apply to every point on the line.

Check your work

Identify the missing perpendicular requirement.

Exit question10

A line is perpendicular to AB but crosses it away from its midpoint. Can it be the perpendicular bisector of AB?

No; perpendicular lines cannot intersect.
No; it fails the midpoint requirement.
Yes; perpendicularity alone is enough.
Yes; a segment has many different midpoints.

Check the point where the lines meet.
A perpendicular bisector must divide AB equally.
A crossing away from the midpoint fails that condition.

Perpendicularity supplies one definition requirement.
The intersection must divide AB into two equal halves as well.

Check your work

Explain why the midpoint condition is indispensable.

Further practice11

M is the midpoint of AB and AB=7.6 cm. Find AM.

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

Recall the definition of midpoint.
The whole length consists of two equal halves.
Divide 7.6 by 2.

The midpoint divides AB into two equal segments.
7.6÷2=3.8, so AM=3.8 cm.

Check your work

Use the correct midpoint or perpendicular-bisector property and retain cm.

Further practice12

P lies on the perpendicular bisector of AB. If PA=11.5 cm, find PB and state the property.

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

Identify the segment whose perpendicular bisector contains P.
Compare distances to the two endpoints, not to the line itself.
Use PB=PA=11.5.

The point is on the perpendicular bisector, so PA=PB.
Substitute PA=11.5 cm to obtain PB=11.5 cm.

Check your work

Use the correct midpoint or perpendicular-bisector property and retain cm.

Further practice13

If A≠B and PA=PB, must P be inside some specified angle before the perpendicular-bisector converse can be used? Contrast this with the angle-bisector converse.

Ask whether the objects are two points or two angle sides.
The perpendicular bisector is a full line.
Do not transfer the angle-interior hypothesis to an unrelated segment theorem.

PA and PB are distances to two endpoints of a segment.
Their equality locates P on the full perpendicular-bisector line on either side of AB.
Distances to angle sides involve a different theorem and its stated interior restriction.

Check your work

Distinguish the two geometric situations and their domains.

Further practice14

To prove PA=PB from M being the midpoint and PM⊥AB, a pupil writes “PA=PB, therefore HL gives congruence.” Identify the circular step and give a valid criterion.

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

Circle the required conclusion in the question.
Check whether that fact has already been used as a premise.
Use two sides and their included right angle instead.

The pupil has assumed the desired equality of the hypotenuses.
The given facts instead supply equal half-segments and two included right angles.
Add common PM and apply SAS, then deduce PA=PB.

Check your work

Identify the assumed conclusion and replace it with justified SAS data.

Extension15

A,B,C are noncollinear. P lies on the perpendicular bisectors of AB and BC. Prove that P also lies on the perpendicular bisector of AC.

Write one equality from each known perpendicular bisector.
Use PB to link PA with PC.
Equal distances to A and C identify AC’s perpendicular bisector.

The property for AB gives PA=PB.
The property for BC gives PB=PC.
Thus PA=PC, and A≠C because the vertices are noncollinear.
Use the endpoint-distance converse for AC.

Check your work

Use two forward properties followed by the correct converse.
Accept another complete proof if every condition and deduction is justified.

Extension16

M is the midpoint of AB. If P=M, the usual SAS triangles collapse. Is the equality PA=PB still true? Justify it without congruent triangles.

Use the definition of midpoint directly.
With P=M, PA is just MA.
Similarly PB=MB=MA.

Replace P by M in the two lengths.
The midpoint definition already gives MA=MB.
No congruence criterion is needed for this special case.

Check your work

Handle the degenerate case directly rather than invoking SAS on collapsed triangles.

Preparation and assessment notes

Prerequisites

Midpoints, perpendicular lines, SAS and SSS, and adjacent angles on a straight line. In triangle proofs, P is off AB; the midpoint case is discussed separately.

Look for in students’ work

For the forward proof use SAS, not HL with the target hypotenuses assumed equal. In the SSS converse, omit the right-angle mark until equal adjacent angles imply 90°. Accept another justified proof.

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

All reference answers

Worked example1

M is the midpoint of AB. P is off AB and PM⊥AB. Prove PA=PB.

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

Reference answer

△PMA≅△PMB by SAS, so PA=PB.

AM=BM because M is the midpoint of AB.
∠PMA=∠PMB=90° because PM⊥AB.
PM=PM is common, so △PMA≅△PMB by SAS.
Thus PA=PB, by corresponding sides of congruent triangles.

Use the midpoint and perpendicular conditions separately.
State the common side and justify the congruence criterion.
Deduce the target without assuming equal endpoint distances first.
Accept another complete proof if every condition and deduction is justified.

Guided attempt2

H is the midpoint of CD, Q is off CD and QH⊥CD. Complete a proof that QC=QD.

CH=___ because H is ___.
∠QHC=∠___=___ because ___.
___ is common; the criterion is ___.
Therefore QC=QD by ___.

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

Reference answer

CH=DH, both included angles at H are 90°, and QH is common; SAS gives QC=QD.

CH=DH because H is the midpoint of CD.
∠QHC=∠QHD=90° because QH⊥CD.
QH=QH is common, so △QHC≅△QHD by SAS.
Thus QC=QD, by corresponding sides of congruent triangles.

Use the midpoint and perpendicular conditions separately.
State the common side and justify the congruence criterion.
Deduce the target without assuming equal endpoint distances first.
Accept another complete proof if every condition and deduction is justified.

Prerequisite3

M is the midpoint of AB and AB=10 cm. Find AM.

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

Reference answer

AM=5 cm.

The midpoint divides AB into two equal segments.
10÷2=5, so AM=5 cm.

Use the correct midpoint or perpendicular-bisector property and retain cm.

Prerequisite4

A,M,B are collinear, P is off AB and PM⊥AB. What is ∠PMA?

45°
180°
It must equal the length AM.
90°

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

Reference answer

90°, by the definition of perpendicular lines.

MA is part of the line AB.
PM⊥AB therefore makes ∠PMA a right angle.

Give an angle of 90°, not a length.

Independent practice5

M is the midpoint of AB and AB=18 cm. Find AM.

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

Reference answer

AM=9 cm.

The midpoint divides AB into two equal segments.
18÷2=9, so AM=9 cm.

Use the correct midpoint or perpendicular-bisector property and retain cm.

Independent practice6

P lies on the perpendicular bisector of AB. If PA=8 cm, find PB and state the property.

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

Reference answer

PB=8 cm: a point on the perpendicular bisector is equidistant from the endpoints.

The point is on the perpendicular bisector, so PA=PB.
Substitute PA=8 cm to obtain PB=8 cm.

Use the correct midpoint or perpendicular-bisector property and retain cm.

Independent practice7

P lies on the perpendicular bisector of AB. PA=3x+1 cm and PB=2x+5 cm. Find x and check both distances.

3x+1=2x+5

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

Reference answer

x=4; PA=PB=13 cm.

The perpendicular-bisector property gives PA=PB, so 3x+1=2x+5.
Subtract 2x and then subtract 1: x=4.
Check: 3×4+1=13 and 2×4+5=13; both are positive.

Justify equal endpoint distances, solve x=4, and verify 13 cm.

Independent practice8

P is off AB and PA=PB. M is the midpoint of AB. Prove PM⊥AB, and conclude that P lies on the perpendicular bisector of AB.

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

Reference answer

△PMA≅△PMB by SSS; the equal adjacent angles at M total 180°, so each is 90°.

PA=PB is given, AM=BM because M is the midpoint, and PM is common.
Therefore △PMA≅△PMB by SSS.
Corresponding angles give ∠PMA=∠PMB.
A,M,B are collinear, so these adjacent angles total 180°; each is 90°.
Thus PM⊥AB through its midpoint, placing P on the perpendicular bisector.

Supply all three SSS side pairs.
Use both equal angles and their straight-angle sum.
Conclude perpendicularity through the midpoint.
Accept another complete proof if every condition and deduction is justified.

Exit question9

A line passes through the midpoint of AB, but it is not stated to be perpendicular to AB. Must every point on it be equidistant from A and B?

No; the perpendicular condition is also needed.
Yes; midpoint information alone is enough.
Yes; all lines through one point coincide.
No line can be perpendicular to AB.

Reference answer

No. A perpendicular bisector requires both midpoint and perpendicularity.

Passing through the midpoint supplies the “bisector” part only.
Without perpendicularity, the equal-distance theorem does not apply to every point on the line.

Identify the missing perpendicular requirement.

Exit question10

A line is perpendicular to AB but crosses it away from its midpoint. Can it be the perpendicular bisector of AB?

No; perpendicular lines cannot intersect.
No; it fails the midpoint requirement.
Yes; perpendicularity alone is enough.
Yes; a segment has many different midpoints.

Reference answer

No. It must also pass through the unique midpoint.

Perpendicularity supplies one definition requirement.
The intersection must divide AB into two equal halves as well.

Explain why the midpoint condition is indispensable.

Further practice11

M is the midpoint of AB and AB=7.6 cm. Find AM.

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

Reference answer

AM=3.8 cm.

The midpoint divides AB into two equal segments.
7.6÷2=3.8, so AM=3.8 cm.

Use the correct midpoint or perpendicular-bisector property and retain cm.

Further practice12

P lies on the perpendicular bisector of AB. If PA=11.5 cm, find PB and state the property.

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

Reference answer

PB=11.5 cm: a point on the perpendicular bisector is equidistant from the endpoints.

The point is on the perpendicular bisector, so PA=PB.
Substitute PA=11.5 cm to obtain PB=11.5 cm.

Use the correct midpoint or perpendicular-bisector property and retain cm.

Further practice13

If A≠B and PA=PB, must P be inside some specified angle before the perpendicular-bisector converse can be used? Contrast this with the angle-bisector converse.

Reference answer

No specified-angle interior restriction is needed for a segment’s perpendicular bisector. The stated internal angle-bisector converse does require P inside that angle.

PA and PB are distances to two endpoints of a segment.
Their equality locates P on the full perpendicular-bisector line on either side of AB.
Distances to angle sides involve a different theorem and its stated interior restriction.

Distinguish the two geometric situations and their domains.

Further practice14

To prove PA=PB from M being the midpoint and PM⊥AB, a pupil writes “PA=PB, therefore HL gives congruence.” Identify the circular step and give a valid criterion.

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

Reference answer

PA=PB was the target, not a given. Use AM=BM, common PM and the included right angles for SAS.

The pupil has assumed the desired equality of the hypotenuses.
The given facts instead supply equal half-segments and two included right angles.
Add common PM and apply SAS, then deduce PA=PB.

Identify the assumed conclusion and replace it with justified SAS data.

Extension15

A,B,C are noncollinear. P lies on the perpendicular bisectors of AB and BC. Prove that P also lies on the perpendicular bisector of AC.

Reference answer

PA=PB and PB=PC, so PA=PC; the converse places P on AC’s perpendicular bisector.

The property for AB gives PA=PB.
The property for BC gives PB=PC.
Thus PA=PC, and A≠C because the vertices are noncollinear.
Use the endpoint-distance converse for AC.

Use two forward properties followed by the correct converse.
Accept another complete proof if every condition and deduction is justified.

Extension16

M is the midpoint of AB. If P=M, the usual SAS triangles collapse. Is the equality PA=PB still true? Justify it without congruent triangles.

Reference answer

Yes: PA=MA and PB=MB, and MA=MB by the definition of midpoint.

Replace P by M in the two lengths.
The midpoint definition already gives MA=MB.
No congruence criterion is needed for this special case.

Handle the degenerate case directly rather than invoking SAS on collapsed triangles.

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. 31–32 · foundation lesson 16

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

Perpendicular bisectors: properties and the SSS converse

Year 8, upper volume · text pp. 31–32 · foundation lesson 16 · 1/2

Name: __________ Class: ______ Date: __________

Worked example1

M is the midpoint of AB. P is off AB and PM⊥AB. Prove PA=PB.

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

Guided attempt2

H is the midpoint of CD, Q is off CD and QH⊥CD. Complete a proof that QC=QD.

CH=___ because H is ___.
∠QHC=∠___=___ because ___.
___ is common; the criterion is ___.
Therefore QC=QD by ___.

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

Prerequisite3

M is the midpoint of AB and AB=10 cm. Find AM.

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

Prerequisite4

A,M,B are collinear, P is off AB and PM⊥AB. What is ∠PMA?

45°
180°
It must equal the length AM.
90°

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

Perpendicular bisectors: properties and the SSS converse

Year 8, upper volume · text pp. 31–32 · foundation lesson 16 · 2/2

Name: __________ Class: ______ Date: __________

Independent practice5

M is the midpoint of AB and AB=18 cm. Find AM.

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

Independent practice6

P lies on the perpendicular bisector of AB. If PA=8 cm, find PB and state the property.

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

Independent practice7

P lies on the perpendicular bisector of AB. PA=3x+1 cm and PB=2x+5 cm. Find x and check both distances.

3x+1=2x+5

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

Independent practice8

P is off AB and PA=PB. M is the midpoint of AB. Prove PM⊥AB, and conclude that P lies on the perpendicular bisector of AB.

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

Exit question9

A line passes through the midpoint of AB, but it is not stated to be perpendicular to AB. Must every point on it be equidistant from A and B?

No; the perpendicular condition is also needed.
Yes; midpoint information alone is enough.
Yes; all lines through one point coincide.
No line can be perpendicular to AB.

Exit question10

A line is perpendicular to AB but crosses it away from its midpoint. Can it be the perpendicular bisector of AB?

No; perpendicular lines cannot intersect.
No; it fails the midpoint requirement.
Yes; perpendicularity alone is enough.
Yes; a segment has many different midpoints.

Perpendicular bisectors: properties and the SSS converse — Further practice

Year 8, upper volume · text pp. 31–32 · foundation lesson 16

Further practice11

M is the midpoint of AB and AB=7.6 cm. Find AM.

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

Further practice12

P lies on the perpendicular bisector of AB. If PA=11.5 cm, find PB and state the property.

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

Further practice13

If A≠B and PA=PB, must P be inside some specified angle before the perpendicular-bisector converse can be used? Contrast this with the angle-bisector converse.

Further practice14

To prove PA=PB from M being the midpoint and PM⊥AB, a pupil writes “PA=PB, therefore HL gives congruence.” Identify the circular step and give a valid criterion.

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

Extension15

A,B,C are noncollinear. P lies on the perpendicular bisectors of AB and BC. Prove that P also lies on the perpendicular bisector of AC.

Extension16

M is the midpoint of AB. If P=M, the usual SAS triangles collapse. Is the equality PA=PB still true? Justify it without congruent triangles.

Perpendicular bisectors: properties and the SSS converse — Teacher answers

Year 8, upper volume · text pp. 31–32 · foundation lesson 16

For the forward proof use SAS, not HL with the target hypotenuses assumed equal. In the SSS converse, omit the right-angle mark until equal adjacent angles imply 90°. Accept another justified proof.

Worked example1

M is the midpoint of AB. P is off AB and PM⊥AB. Prove PA=PB.

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

Reference answer

△PMA≅△PMB by SAS, so PA=PB.

AM=BM because M is the midpoint of AB.
∠PMA=∠PMB=90° because PM⊥AB.
PM=PM is common, so △PMA≅△PMB by SAS.
Thus PA=PB, by corresponding sides of congruent triangles.

Use the midpoint and perpendicular conditions separately.
State the common side and justify the congruence criterion.
Deduce the target without assuming equal endpoint distances first.
Accept another complete proof if every condition and deduction is justified.

Guided attempt2

H is the midpoint of CD, Q is off CD and QH⊥CD. Complete a proof that QC=QD.

CH=___ because H is ___.
∠QHC=∠___=___ because ___.
___ is common; the criterion is ___.
Therefore QC=QD by ___.

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

Reference answer

CH=DH, both included angles at H are 90°, and QH is common; SAS gives QC=QD.

CH=DH because H is the midpoint of CD.
∠QHC=∠QHD=90° because QH⊥CD.
QH=QH is common, so △QHC≅△QHD by SAS.
Thus QC=QD, by corresponding sides of congruent triangles.

Use the midpoint and perpendicular conditions separately.
State the common side and justify the congruence criterion.
Deduce the target without assuming equal endpoint distances first.
Accept another complete proof if every condition and deduction is justified.

Prerequisite3

M is the midpoint of AB and AB=10 cm. Find AM.

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

Reference answer

AM=5 cm.

The midpoint divides AB into two equal segments.
10÷2=5, so AM=5 cm.

Use the correct midpoint or perpendicular-bisector property and retain cm.

Prerequisite4

A,M,B are collinear, P is off AB and PM⊥AB. What is ∠PMA?

45°
180°
It must equal the length AM.
90°

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

Reference answer

90°, by the definition of perpendicular lines.

MA is part of the line AB.
PM⊥AB therefore makes ∠PMA a right angle.

Give an angle of 90°, not a length.

Independent practice5

M is the midpoint of AB and AB=18 cm. Find AM.

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

Reference answer

AM=9 cm.

The midpoint divides AB into two equal segments.
18÷2=9, so AM=9 cm.

Use the correct midpoint or perpendicular-bisector property and retain cm.

Independent practice6

P lies on the perpendicular bisector of AB. If PA=8 cm, find PB and state the property.

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

Reference answer

PB=8 cm: a point on the perpendicular bisector is equidistant from the endpoints.

The point is on the perpendicular bisector, so PA=PB.
Substitute PA=8 cm to obtain PB=8 cm.

Use the correct midpoint or perpendicular-bisector property and retain cm.

Independent practice7

P lies on the perpendicular bisector of AB. PA=3x+1 cm and PB=2x+5 cm. Find x and check both distances.

3x+1=2x+5

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

Reference answer

x=4; PA=PB=13 cm.

The perpendicular-bisector property gives PA=PB, so 3x+1=2x+5.
Subtract 2x and then subtract 1: x=4.
Check: 3×4+1=13 and 2×4+5=13; both are positive.

Justify equal endpoint distances, solve x=4, and verify 13 cm.

Independent practice8

P is off AB and PA=PB. M is the midpoint of AB. Prove PM⊥AB, and conclude that P lies on the perpendicular bisector of AB.

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

Reference answer

△PMA≅△PMB by SSS; the equal adjacent angles at M total 180°, so each is 90°.

PA=PB is given, AM=BM because M is the midpoint, and PM is common.
Therefore △PMA≅△PMB by SSS.
Corresponding angles give ∠PMA=∠PMB.
A,M,B are collinear, so these adjacent angles total 180°; each is 90°.
Thus PM⊥AB through its midpoint, placing P on the perpendicular bisector.

Supply all three SSS side pairs.
Use both equal angles and their straight-angle sum.
Conclude perpendicularity through the midpoint.
Accept another complete proof if every condition and deduction is justified.

Exit question9

A line passes through the midpoint of AB, but it is not stated to be perpendicular to AB. Must every point on it be equidistant from A and B?

No; the perpendicular condition is also needed.
Yes; midpoint information alone is enough.
Yes; all lines through one point coincide.
No line can be perpendicular to AB.

Reference answer

No. A perpendicular bisector requires both midpoint and perpendicularity.

Passing through the midpoint supplies the “bisector” part only.
Without perpendicularity, the equal-distance theorem does not apply to every point on the line.

Identify the missing perpendicular requirement.

Exit question10

A line is perpendicular to AB but crosses it away from its midpoint. Can it be the perpendicular bisector of AB?

No; perpendicular lines cannot intersect.
No; it fails the midpoint requirement.
Yes; perpendicularity alone is enough.
Yes; a segment has many different midpoints.

Reference answer

No. It must also pass through the unique midpoint.

Perpendicularity supplies one definition requirement.
The intersection must divide AB into two equal halves as well.

Explain why the midpoint condition is indispensable.

Further practice11

M is the midpoint of AB and AB=7.6 cm. Find AM.

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

Reference answer

AM=3.8 cm.

The midpoint divides AB into two equal segments.
7.6÷2=3.8, so AM=3.8 cm.

Use the correct midpoint or perpendicular-bisector property and retain cm.

Further practice12

P lies on the perpendicular bisector of AB. If PA=11.5 cm, find PB and state the property.

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

Reference answer

PB=11.5 cm: a point on the perpendicular bisector is equidistant from the endpoints.

The point is on the perpendicular bisector, so PA=PB.
Substitute PA=11.5 cm to obtain PB=11.5 cm.

Use the correct midpoint or perpendicular-bisector property and retain cm.

Further practice13

If A≠B and PA=PB, must P be inside some specified angle before the perpendicular-bisector converse can be used? Contrast this with the angle-bisector converse.

Reference answer

No specified-angle interior restriction is needed for a segment’s perpendicular bisector. The stated internal angle-bisector converse does require P inside that angle.

PA and PB are distances to two endpoints of a segment.
Their equality locates P on the full perpendicular-bisector line on either side of AB.
Distances to angle sides involve a different theorem and its stated interior restriction.

Distinguish the two geometric situations and their domains.

Further practice14

To prove PA=PB from M being the midpoint and PM⊥AB, a pupil writes “PA=PB, therefore HL gives congruence.” Identify the circular step and give a valid criterion.

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

Reference answer

PA=PB was the target, not a given. Use AM=BM, common PM and the included right angles for SAS.

The pupil has assumed the desired equality of the hypotenuses.
The given facts instead supply equal half-segments and two included right angles.
Add common PM and apply SAS, then deduce PA=PB.

Identify the assumed conclusion and replace it with justified SAS data.

Extension15

A,B,C are noncollinear. P lies on the perpendicular bisectors of AB and BC. Prove that P also lies on the perpendicular bisector of AC.

Reference answer

PA=PB and PB=PC, so PA=PC; the converse places P on AC’s perpendicular bisector.

The property for AB gives PA=PB.
The property for BC gives PB=PC.
Thus PA=PC, and A≠C because the vertices are noncollinear.
Use the endpoint-distance converse for AC.

Use two forward properties followed by the correct converse.
Accept another complete proof if every condition and deduction is justified.

Extension16

M is the midpoint of AB. If P=M, the usual SAS triangles collapse. Is the equality PA=PB still true? Justify it without congruent triangles.

Reference answer

Yes: PA=MA and PB=MB, and MA=MB by the definition of midpoint.

Replace P by M in the two lengths.
The midpoint definition already gives MA=MB.
No congruence criterion is needed for this special case.

Handle the degenerate case directly rather than invoking SAS on collapsed triangles.