Use triangle angle sums and exterior angles
Explain the triangle sum using an auxiliary parallel, then apply it and the correctly marked exterior-angle property with reasons.
Exam scope and assessment notes
Foundation and Higher share all four section4.1 requirements. The Higher table says See Foundation Tier; there are no additional H4.1 rows. These six original lessons supply selected angle types, local point/straight/intersection rules, marked parallel-transversal relations, triangle sums and exterior angles, isosceles/equilateral properties and right/mixed triangle angles. F:4.1B requires both the local-line and parallel-line lessons; F:4.1D requires both special-triangle and right-triangle lessons with substantive teaching, a worked example and every Guided, Independent and Transfer stage. Six lessons and24 selected seed cases are StudyForA’s design, not a syllabus-prescribed count. Draw or copy each explicitly labelled configuration, retain the selected reflex or nonreflex exterior arc and mark only given or justified collinearity, parallel arrows, equality ticks and right squares. Diagrams need not be to scale; do not measure them unless instructed. State geometric reasons before equations, substitute back and reject impossible or insufficient data. The triangle-sum explanation uses an auxiliary parallel through a vertex with transported alternate angles and a straight-angle partition; the exterior relationship requires an actual straight side extension. Equilateral is also isosceles under the inclusive at-least-two convention, and right-isosceles overlap remains valid. A fixed reference picture and a correct number do not establish that the learner drew or justified the configuration. Written work is not automatically marked. Attempt acknowledgements and revealed references do not award marks or establish mastery. These structural links are not exhaustive subskill approval; specialist approval and formal assessment validation remain unestablished. Section4.5 instrument constructions and4.4 measurement/bearings remain separate, as does the full H4.7 reasoning scope across lines, polygons and circles. No blanket formal-proof waiver is imported from unrelated circle-theorem notes. The wider Maths A course remains incomplete. Publisher book navigation is separate evidence: Book1 Shape and Space1 has literal F4.1B/C/D chapter tags, not F4.1A; Hub18 literally tags B/D, not A/C. Map pages60–99 and proof chapter start60 agree here, with the next proof chapter at100; do not invent a pagination conflict or claim final-edition approval. Only selected public uncorrected proof pages60–65 were previously inspected. Full final-book and Teaching Hub body/pagination alignment remains unverified; chapter links confer no full-book or modular teaching credit.
Before you begin
- Use corresponding or alternate angles for stated parallel lines.
- Use straight-angle sums and solve simple linear equations.
- Name an angle with its vertex letter in the middle.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Triangle ABC has A=52° and B=63°. What is its interior C?
What you will learn
- Explain why the three interior angles of a nondegenerate plane triangle sum to180°.
- Identify a stated straight side extension and distinguish its nonreflex exterior from adjacent and remote interiors.
- Solve and check angle expressions using both the triangle sum and exterior property.
Draw a nondegenerate labelled triangle ABC with A above the line BC. Extend BC through C to D, so B,C,D are collinear in that order. Mark the three interior arcs and the nonreflex exterior∠ACD between CA and CD separately. Given ∠BAC=48° and∠ABC=67°, find∠ACB and∠ACD with two checks for the exterior. Also draw a line E–A–F parallel to BC, with AE toward the B side and AF toward the C side, to explain the triangle-sum property.
Copy and annotate the labelled figures on paper. Retain every given, chosen arc, straight extension and geometric reason. Exact givens decide deductions; these screen diagrams must not be measured unless a task explicitly asks. Unknown values and classifications stay hidden until you attempt your reasoning.
Triangle ABC with a true BC extension
- ∠BAC: given 48°
- ∠ABC: given 67°
- ∠ACB: unknown
- ∠ACD: unknown
Separate feasibility check: this does not change any fixed exercise or mark your work.
Explain the triangle sum with an auxiliary parallel
Draw triangle ABC with BC as a base and A on one side of its line. Through A draw the straight line EAF parallel to BC, with E on the B side and F on the C side. The alternate angles∠EAB and∠ABC are equal, as are∠CAF and∠ACB. Mark the parallel arrows and these transported angles on the diagram.
At A, the angles∠EAB,∠BAC,∠CAF are consecutive along straight EAF, so they add to180°. Replacing the transported angles by B and C gives A+B+C=180°. This geometric argument applies to the nondegenerate triangle; checking or measuring a single example would not establish the general property.
An exterior needs an actual straight extension
Extend BC through C to D, with B,C,D collinear in that order. The selected nonreflex exterior∠ACD is between CA and CD. It shares CA with the adjacent interior∠ACB, and their other rays CB and CD are opposite, so the interior and exterior add to180°. Mark this arc instead of a reflex sweep.
The triangle sum gives A+B=180−C, and the straight extension gives exterior=180−C. Therefore the exterior equals A+B, the two remote interior angles. It is not normally equal to its adjacent interior. If the extra ray is not a stated straight side extension, this particular exterior relation cannot simply be assumed.
Solve, annotate and check feasibility
Use the triangle-sum or exterior relationship to form an equation, naming its geometric reason. Solve for the parameter, substitute into the original labels and mark the actual degree measures at the named vertices. A nonreflex exterior can be checked in two ways: as the remote-angle sum and as180° minus the adjacent interior.
For a nondegenerate triangle every interior lies strictly between0° and180°, and all three sum exactly180°. A zero-angle or inconsistent sum is not repaired by drawing a nicer-looking triangle. State that the data are impossible or insufficient when appropriate. Give the required geometric explanation; the separate construction and broader reasoning topics are not automatically completed by these sketches.
Sources for this lesson
Pause and explain
In triangle ABC, BC extends through C to D and the nonreflex exterior∠ACD is marked. Which pair sums to that exterior?
Worked example
Draw a nondegenerate labelled triangle ABC with A above the line BC. Extend BC through C to D, so B,C,D are collinear in that order. Mark the three interior arcs and the nonreflex exterior∠ACD between CA and CD separately. Given ∠BAC=48° and∠ABC=67°, find∠ACB and∠ACD with two checks for the exterior. Also draw a line E–A–F parallel to BC, with AE toward the B side and AF toward the C side, to explain the triangle-sum property.
Show the worked solution
- For the triangle-sum argument, use the auxiliary EAF line through A parallel to BC. Alternate interior angles give∠EAB=∠ABC and∠CAF=∠ACB. The three adjacent angles∠EAB,∠BAC,∠CAF fill straight∠EAF=180°, so A+B+C=180°. This explains the property for a nondegenerate triangle rather than inferring it from one measurement.
- Apply the property: C=180−48−67=65°. All interior angles48°,67°,65° are positive and sum to180°. Mark C at the triangle’s interior, not on the side extension.
- The adjacent interior/exterior pair at C fills a straight angle, so∠ACD=180−65=115°. As a separate check, the exterior equals the two remote interiors:48+67=115°. The selected exterior is the nonreflex arc between CA and CD.
Answer Interiors A=48°,B=67°,C=65°; nonreflex exterior∠ACD=115°. Checks:115+65=180 and115=48+67.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
Draw a nondegenerate labelled triangle ABC with A above the line BC. Extend BC through C to D, so B,C,D are collinear in that order. Mark the three interior arcs and the nonreflex exterior∠ACD between CA and CD separately. Given A=(2x+7)°, B=(3x−2)° and C=70°, find x, every interior angle and the marked nonreflex exterior at C. State the geometric reason for the equation.
Give me a hint
Use the labelled triangle and actual straight extension to identify the adjacent interior and two remote interiors before writing the equation.
Compare my reasoning
- The three interior angles sum to180°, so(2x+7)+(3x−2)+70=180. Thus5x+75=180 and x=21.
- Substitute: A=49°, B=61° and C=70°. These positive interior angles total180°. Keep the algebraic parameter21 distinct from any of these angle measures.
- Because BC really extends through C to D, exterior∠ACD=180−70=110°. The remote-angle check gives49+61=110°. Annotate both reasons on the actual diagram.
x=21; interiors49°,61°,70°; exterior at C110°.
Look for these in your work
- Draw the labelled nondegenerate triangle, its extension and the correct selected exterior arc.
- Name the triangle-sum or remote-exterior reason before solving.
- Recover all angles and check positivity, interior total and the supplementary exterior pair.
Draw a nondegenerate labelled triangle ABC with A above the line BC. Extend BC through C to D, so B,C,D are collinear in that order. Mark the three interior arcs and the nonreflex exterior∠ACD between CA and CD separately. Given∠ACD=(4x+6)°, A=(x+16)° and B=(2x+14)°, find x, all interior angles and the exterior. Identify the two remote interior angles on the actual extension diagram.
Give me a hint
Use the labelled triangle and actual straight extension to identify the adjacent interior and two remote interiors before writing the equation.
Compare my reasoning
- The chosen nonreflex exterior equals its two remote interior angles, so4x+6=(x+16)+(2x+14). Hence4x+6=3x+30 and x=24.
- Substitute to obtain A=40°, B=62° and exterior∠ACD=102°. Its adjacent interior is C=180−102=78°.
- Verify40+62+78=180 and40+62=102. Every interior is positive, and78+102=180 at the actual side extension. The two remote angles are A and B, not the adjacent interior C.
x=24; interiors40°,62°,78°; exterior at C102°.
Look for these in your work
- Draw the labelled nondegenerate triangle, its extension and the correct selected exterior arc.
- Name the triangle-sum or remote-exterior reason before solving.
- Recover all angles and check positivity, interior total and the supplementary exterior pair.
Draw a nondegenerate labelled triangle ABC with A above the line BC. Extend BC through C to D, so B,C,D are collinear in that order. Mark the three interior arcs and the nonreflex exterior∠ACD between CA and CD separately. The triangle models a fictional support with A=(y+12)°, B=(2y+18)° and nonreflex exterior at C=(4y−5)°. Find y and every angle. Explain why the exterior is not equal to its adjacent interior in this support.
Give me a hint
Use the labelled triangle and actual straight extension to identify the adjacent interior and two remote interiors before writing the equation.
Compare my reasoning
- The marked exterior and two remote interiors give4y−5=(y+12)+(2y+18), so4y−5=3y+30 and y=35.
- The actual remote interiors are A=47° and B=88°, while the exterior is135°. The adjacent interior C=180−135=45°.
- Check47+88+45=180 and47+88=135. The exterior135° and adjacent interior45° are supplementary, not equal here. The support diagram must retain B,C,D collinear; an arbitrary ray from C would not license this exterior-angle deduction.
y=35; interiors47°,88°,45°; exterior at C135°, supplementary to the45° adjacent interior.
Look for these in your work
- Draw the labelled nondegenerate triangle, its extension and the correct selected exterior arc.
- Name the triangle-sum or remote-exterior reason before solving.
- Recover all angles and check positivity, interior total and the supplementary exterior pair.
Common mistakes
- Using a triangle exterior property on an arbitrary extra ray without a straight side extension.
- Adding the adjacent interior to one remote angle instead of using both remote interiors.
- Omitting the auxiliary-parallel explanation or accepting nonpositive/inconsistent interior angles.
What can you explain now?
A proposed triangle ABC has exact interiors A=0°,B=80°,C=100°. Does the180° total make it a valid nondegenerate triangle?
Compare with the explanation
No. The zero interior makes it degenerate.
A valid nondegenerate triangle needs each interior strictly positive as well as a180° total. Do not draw an ordinary triangle and silently change the zero label.
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Make it stick
Explain the triangle sum with a labelled auxiliary parallel. Then derive the remote exterior-angle property and state the diagram and positivity conditions it requires.
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.