PAPER · REASONING · PRACTICE

Isosceles, equilateral and special-triangle practice

Apply properties and converses with conditions; use 30° right triangles and an optional reflection shortest-path extension.

40 minutes · Year 8, upper volume · text pp. 39–50 · foundation lessons 20–25

Textbook reference: Year 8, upper · Chapter 13: reflection symmetry · lessons 20–25

Chapter core practice: use after the referenced textbook lessons, or for chapter review. See the upper-volume plan

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

Fold an isosceles triangle from its apex. Compare the altitude, median and angle bisector, then ask which hypotheses make them coincide.

Organise the key idea

Equal sides and equal opposite angles are converse tests. Equilateral triangles have three 60° angles. In a right triangle, the leg opposite 30° is half the hypotenuse.

Open the original chapter

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

Worked example1

AB=AC and ∠A=40°. Find each base angle.

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

Identify the known quantities and the required quantity.
Write the relevant equality before calculating.
Substitute carefully, retain signs and units, and check the result.

Equal sides give equal opposite base angles: ∠B=∠C.
2∠B=180°−40°, so each base angle is 70°.

Check your work

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Guided attempt2

In △ABC, ∠B=∠C. Construct the internal bisector AD with D on BC. Complete a congruence proof of AB=AC.

∠ABD=___; ∠BAD=___; AD=___.
△ABD≅△___ by ___; hence ___=___.

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

Compare △ABD and △ACD.
Use the given base angles, the constructed apex angles and common AD.
AAS gives the side equality after congruence.

∠ABD=∠ACD because D lies on BC and ∠B=∠C.
∠BAD=∠CAD by the constructed angle bisector; AD=AD is common.
AAS gives △ABD≅△ACD, and corresponding sides give AB=AC.

Check your work

Give two angle pairs and common AD before AAS.
Do not assume AB=AC at the start.

Prerequisite3

Two equal base angles are each 65°. Find the vertex angle.

Identify the known quantities and the required quantity.
Write the relevant equality before calculating.
Substitute carefully, retain signs and units, and check the result.

The three angles total 180°.
180−2×65=50°.

Check your work

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Prerequisite4

Find each interior angle of an equilateral triangle.

Identify the known quantities and the required quantity.
Write the relevant equality before calculating.
Substitute carefully, retain signs and units, and check the result.

All three angles are equal because all three sides are equal.
Each angle is 180°/3=60°.

Check your work

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Independent practice5

AB=AC and ∠A=52°. Find each base angle.

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

Identify the known quantities and the required quantity.
Write the relevant equality before calculating.
Substitute carefully, retain signs and units, and check the result.

Equal sides give equal opposite base angles: ∠B=∠C.
2∠B=180°−52°, so each base angle is 64°.

Check your work

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Independent practice6

If ∠ABC=∠ACB, which side equality follows?

All sides are zero.
AB=AC
AB=BC
AC=BC

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

Equal angles have equal opposite sides.
The opposite sides to B and C are AC and AB.
Compare with: AB=AC

Equal angles have equal opposite sides.
The opposite sides to B and C are AC and AB.

Check your work

Choose the statement justified by the stated conditions and explain why.

Independent practice7

AB=AC and ∠A=60°. Prove that △ABC is equilateral; give the angle and side reasons.

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

Use the equal-side property first.
Subtract 60° from 180° and split the remainder equally.
Use the equal-angle converse to obtain the third equal side.

AB=AC gives ∠B=∠C.
Each base angle is (180−60)/2=60°.
All three angles are equal, so their opposite sides are equal; the triangle is equilateral.

Check your work

State all hypotheses and justify each deduction.
Do not assume the conclusion. Accept another complete justified proof.

Independent practice8

∠C=90°, ∠A=30° and AB=10 cm. Find BC.

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

Identify the known quantities and the required quantity.
Write the relevant equality before calculating.
Substitute carefully, retain signs and units, and check the result.

BC is opposite 30° and AB is the hypotenuse.
BC=AB/2=10/2=5 cm.

Check your work

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Exit question9

An equilateral triangle has perimeter 15 cm. Find AB.

Identify the known quantities and the required quantity.
Write the relevant equality before calculating.
Substitute carefully, retain signs and units, and check the result.

All three sides have equal length.
AB=15/3=5 cm.

Check your work

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Exit question10

In a right triangle, the leg opposite 30° is 4 cm. Find the hypotenuse AB.

Identify the known quantities and the required quantity.
Write the relevant equality before calculating.
Substitute carefully, retain signs and units, and check the result.

The stated leg is half the hypotenuse.
AB=2×4=8 cm.

Check your work

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Further practice11

AB=AC and ∠A=84°. Find each base angle.

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

Identify the known quantities and the required quantity.
Write the relevant equality before calculating.
Substitute carefully, retain signs and units, and check the result.

Equal sides give equal opposite base angles: ∠B=∠C.
2∠B=180°−84°, so each base angle is 48°.

Check your work

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Further practice12

An equilateral triangle has side 7 cm. Find its perimeter.

Identify the known quantities and the required quantity.
Write the relevant equality before calculating.
Substitute carefully, retain signs and units, and check the result.

The three sides are 7,7,7 cm.
The perimeter is 3×7=21 cm.

Check your work

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Further practice13

∠C=90°, ∠A=30° and AB=9 cm. Find BC.

Identify the known quantities and the required quantity.
Write the relevant equality before calculating.
Substitute carefully, retain signs and units, and check the result.

BC is the leg opposite 30°.
BC=9/2=4.5 cm.

Check your work

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Further practice14

AB=AC and each base angle is 38°. Find ∠A.

Identify the known quantities and the required quantity.
Write the relevant equality before calculating.
Substitute carefully, retain signs and units, and check the result.

The two equal base angles total 76°.
∠A=180−76=104°.

Check your work

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Extension15

A=(−3,3) and B=(5,3) are on the same side of the x-axis. A′=(−3,−3) is A’s reflection and A′B=10 units. Find the minimum AQ+QB for Q on the axis and explain why.

Use the labelled axes and unit grid.

Reflect one endpoint across the given line.
Replace the broken path AQ+QB by A′Q+QB.
A straight segment is shortest; use the supplied length 10 rather than measuring.

For Q on the axis, AQ=A′Q by reflection.
AQ+QB=A′Q+QB≥A′B=10 by the triangle inequality.
Equality occurs on the straight segment A′B, at its axis crossing Q=(1,0).

Check your work

Use reflection and the triangle inequality, including the equality location.

Extension16

AB=AC and M is the midpoint of BC. Prove that AM is both an altitude and the internal angle bisector.

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

Compare the two triangles on either side of AM.
Use the equal sides, the equal half-base and common AM for SSS.
Use corresponding apex angles and the equal adjacent angles on BC.

AB=AC is given, BM=CM follows from the midpoint, and AM is common.
Thus △ABM≅△ACM by SSS.
∠BAM=∠MAC; also ∠AMB=∠AMC and they total 180°, so each is 90°.

Check your work

State all hypotheses and justify each deduction.
Do not assume the conclusion. Accept another complete justified proof.

Preparation and assessment notes

Prerequisites

Triangle angle sums, congruence, internal bisectors and perpendicular bisectors.

Look for in students’ work

Use the worked model and guided attempt selectively, then the eight core questions. Keep further practice and extensions optional. Ask for a reason and an original-equation or expansion check.

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

Chapter-core practice after the mapped textbook lessons; these tasks do not replace every first-teaching period.

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. 39–50 · foundation lessons 20–25

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

Isosceles, equilateral and special-triangle practice

Year 8, upper volume · text pp. 39–50 · foundation lessons 20–25 · 1/2

Name: __________ Class: ______ Date: __________

Worked example1

AB=AC and ∠A=40°. Find each base angle.

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

Guided attempt2

In △ABC, ∠B=∠C. Construct the internal bisector AD with D on BC. Complete a congruence proof of AB=AC.

∠ABD=___; ∠BAD=___; AD=___.
△ABD≅△___ by ___; hence ___=___.

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

Prerequisite3

Two equal base angles are each 65°. Find the vertex angle.

Prerequisite4

Find each interior angle of an equilateral triangle.

Isosceles, equilateral and special-triangle practice

Year 8, upper volume · text pp. 39–50 · foundation lessons 20–25 · 2/2

Name: __________ Class: ______ Date: __________

Independent practice5

AB=AC and ∠A=52°. Find each base angle.

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

Independent practice6

If ∠ABC=∠ACB, which side equality follows?

All sides are zero.
AB=AC
AB=BC
AC=BC

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

Independent practice7

AB=AC and ∠A=60°. Prove that △ABC is equilateral; give the angle and side reasons.

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

Independent practice8

∠C=90°, ∠A=30° and AB=10 cm. Find BC.

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

Exit question9

An equilateral triangle has perimeter 15 cm. Find AB.

Exit question10

In a right triangle, the leg opposite 30° is 4 cm. Find the hypotenuse AB.

Isosceles, equilateral and special-triangle practice — Further practice

Year 8, upper volume · text pp. 39–50 · foundation lessons 20–25

Further practice11

AB=AC and ∠A=84°. Find each base angle.

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

Further practice12

An equilateral triangle has side 7 cm. Find its perimeter.

Further practice13

∠C=90°, ∠A=30° and AB=9 cm. Find BC.

Further practice14

AB=AC and each base angle is 38°. Find ∠A.

Extension15

A=(−3,3) and B=(5,3) are on the same side of the x-axis. A′=(−3,−3) is A’s reflection and A′B=10 units. Find the minimum AQ+QB for Q on the axis and explain why.

Use the labelled axes and unit grid.

Extension16

AB=AC and M is the midpoint of BC. Prove that AM is both an altitude and the internal angle bisector.

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

Isosceles, equilateral and special-triangle practice — Teacher answers

Year 8, upper volume · text pp. 39–50 · foundation lessons 20–25

Use the worked model and guided attempt selectively, then the eight core questions. Keep further practice and extensions optional. Ask for a reason and an original-equation or expansion check.

Worked example1

AB=AC and ∠A=40°. Find each base angle.

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

Reference answer

70 °.

(180-40)/2

Equal sides give equal opposite base angles: ∠B=∠C.
2∠B=180°−40°, so each base angle is 70°.

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Guided attempt2

In △ABC, ∠B=∠C. Construct the internal bisector AD with D on BC. Complete a congruence proof of AB=AC.

∠ABD=___; ∠BAD=___; AD=___.
△ABD≅△___ by ___; hence ___=___.

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

Reference answer

△ABD≅△ACD by AAS, so AB=AC.

∠ABD=∠ACD because D lies on BC and ∠B=∠C.
∠BAD=∠CAD by the constructed angle bisector; AD=AD is common.
AAS gives △ABD≅△ACD, and corresponding sides give AB=AC.

Give two angle pairs and common AD before AAS.
Do not assume AB=AC at the start.

Prerequisite3

Two equal base angles are each 65°. Find the vertex angle.

Reference answer

50 °.

180-2*65

The three angles total 180°.
180−2×65=50°.

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Prerequisite4

Find each interior angle of an equilateral triangle.

Reference answer

60 °.

180/3

All three angles are equal because all three sides are equal.
Each angle is 180°/3=60°.

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Independent practice5

AB=AC and ∠A=52°. Find each base angle.

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

Reference answer

64 °.

(180-52)/2

Equal sides give equal opposite base angles: ∠B=∠C.
2∠B=180°−52°, so each base angle is 64°.

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Independent practice6

If ∠ABC=∠ACB, which side equality follows?

All sides are zero.
AB=AC
AB=BC
AC=BC

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

Reference answer

AB=AC

Equal angles have equal opposite sides.
The opposite sides to B and C are AC and AB.

Choose the statement justified by the stated conditions and explain why.

Independent practice7

AB=AC and ∠A=60°. Prove that △ABC is equilateral; give the angle and side reasons.

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

Reference answer

∠B=∠C=60°; equal opposite angles give AB=AC=BC.

AB=AC gives ∠B=∠C.
Each base angle is (180−60)/2=60°.
All three angles are equal, so their opposite sides are equal; the triangle is equilateral.

State all hypotheses and justify each deduction.
Do not assume the conclusion. Accept another complete justified proof.

Independent practice8

∠C=90°, ∠A=30° and AB=10 cm. Find BC.

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

Reference answer

5 cm.

10/2

BC is opposite 30° and AB is the hypotenuse.
BC=AB/2=10/2=5 cm.

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Exit question9

An equilateral triangle has perimeter 15 cm. Find AB.

Reference answer

5 cm.

15/3

All three sides have equal length.
AB=15/3=5 cm.

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Exit question10

In a right triangle, the leg opposite 30° is 4 cm. Find the hypotenuse AB.

Reference answer

8 cm.

2*4

The stated leg is half the hypotenuse.
AB=2×4=8 cm.

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Further practice11

AB=AC and ∠A=84°. Find each base angle.

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

Reference answer

48 °.

(180-84)/2

Equal sides give equal opposite base angles: ∠B=∠C.
2∠B=180°−84°, so each base angle is 48°.

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Further practice12

An equilateral triangle has side 7 cm. Find its perimeter.

Reference answer

21 cm.

3*7

The three sides are 7,7,7 cm.
The perimeter is 3×7=21 cm.

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Further practice13

∠C=90°, ∠A=30° and AB=9 cm. Find BC.

Reference answer

4.5 cm.

9/2

BC is the leg opposite 30°.
BC=9/2=4.5 cm.

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Further practice14

AB=AC and each base angle is 38°. Find ∠A.

Reference answer

104 °.

180-2*38

The two equal base angles total 76°.
∠A=180−76=104°.

Use the stated relationship and show the calculation.
State the result with its unit where appropriate.

Extension15

A=(−3,3) and B=(5,3) are on the same side of the x-axis. A′=(−3,−3) is A’s reflection and A′B=10 units. Find the minimum AQ+QB for Q on the axis and explain why.

Use the labelled axes and unit grid.

Reference answer

10 units, attained at Q=(1,0), the intersection of A′B with the axis.

For Q on the axis, AQ=A′Q by reflection.
AQ+QB=A′Q+QB≥A′B=10 by the triangle inequality.
Equality occurs on the straight segment A′B, at its axis crossing Q=(1,0).

Use reflection and the triangle inequality, including the equality location.

Use the labelled axes and unit grid.

Extension16

AB=AC and M is the midpoint of BC. Prove that AM is both an altitude and the internal angle bisector.

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

Reference answer

△ABM≅△ACM by SSS; the equal adjacent angles at M are 90°, and the two apex angles are equal.

AB=AC is given, BM=CM follows from the midpoint, and AM is common.
Thus △ABM≅△ACM by SSS.
∠BAM=∠MAC; also ∠AMB=∠AMC and they total 180°, so each is 90°.

State all hypotheses and justify each deduction.
Do not assume the conclusion. Accept another complete justified proof.