To bisect a given rectilinear angle.
Given any angle, construct a line that divides it into two equal parts.
Before You Read
Grab a compass and straightedge. Someone draws an angle on your paper—any angle. Can you split it exactly in half without a protractor? No measuring allowed, just the compass and straight edge. Think about what tools you've built so far: equilateral triangles, length copying, and SSS congruence.
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All Foundations
Definitions (23)
- D1 — Point
- D2 — Line
- D3 — Ends of a Line
- D4 — Straight Line
- D5 — Surface
- D6 — Edges of a Surface
- D7 — Plane Surface
- D8 — Plane Angle●
- D9 — Rectilinear Angle●
- D10 — Right Angle & Perpendicular
- D11 — Obtuse Angle
- D12 — Acute Angle
- D13 — Boundary
- D14 — Figure
- D15 — Circle
- D16 — Center of a Circle
- D17 — Diameter
- D18 — Semicircle
- D19 — Rectilinear Figures
- D20 — Types of Triangles (by sides)●
- D21 — Types of Triangles (by angles)
- D22 — Quadrilaterals
- D23 — Parallel Lines
Postulates (5)
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All Propositions
Basic Constructions (5)
- Prop 1 — To construct an equilateral triangle on a given fi…
- Prop 2 — To place a straight line equal to a given straight…
- Prop 3 — To cut off from the greater of two given unequal s…
- Prop 4 — If two triangles have two sides equal to two sides…
- Prop 5 — In isosceles triangles the angles at the base equa…
Triangle Fundamentals (5)
Perpendiculars & Angles (5)
- Prop 11 — To draw a straight line at right angles to a given…
- Prop 12 — To draw a straight line perpendicular to a given i…
- Prop 13 — If a straight line stands on a straight line, then…
- Prop 14 — If with any straight line, and at a point on it, t…
- Prop 15 — If two straight lines cut one another, then they m…
Exterior Angles & Inequalities (5)
- Prop 16 — In any triangle, if one of the sides is produced, …
- Prop 17 — In any triangle the sum of any two angles is less …
- Prop 18 — In any triangle the angle opposite the greater sid…
- Prop 19 — In any triangle the side opposite the greater angl…
- Prop 20 — In any triangle the sum of any two sides is greate…
Interior Triangles & Angle Copying (5)
- Prop 21 — If from the ends of one of the sides of a triangle…
- Prop 22 — To construct a triangle out of three straight line…
- Prop 23 — To construct a rectilinear angle equal to a given …
- Prop 24 — If two triangles have two sides equal to two sides…
- Prop 25 — If two triangles have two sides equal to two sides…
Parallel Lines (5)
- Prop 26 — If two triangles have two angles equal to two angl…
- Prop 27 — If a straight line falling on two straight lines m…
- Prop 28 — If a straight line falling on two straight lines m…
- Prop 29 — A straight line falling on parallel straight lines…
- Prop 30 — Straight lines parallel to the same straight line …
Parallel Constructions & Parallelograms (5)
- Prop 31 — To draw a straight line through a given point para…
- Prop 32 — In any triangle, if one of the sides is produced, …
- Prop 33 — Straight lines which join the ends of equal and pa…
- Prop 34 — In parallelogrammic areas the opposite sides and a…
- Prop 35 — Parallelograms which are on the same base and in t…
Area Theorems (5)
- Prop 36 — Parallelograms which are on equal bases and in the…
- Prop 37 — Triangles which are on the same base and in the sa…
- Prop 38 — Triangles which are on equal bases and in the same…
- Prop 39 — Equal triangles which are on the same base and on …
- Prop 40 — Equal triangles which are on equal bases and on th…
Area Applications (4)
What Euclid Is Doing
Setup: We have ∠BAC and want to find a line from A that splits it exactly in half.
Approach: The strategy is beautifully simple: manufacture two triangles that share the bisector line AF, then prove they're congruent by SSS. Here's how: mark equal distances along both sides of the angle (AD = AE), then build an equilateral triangle on DE to get a point F where DF = EF. Now triangles ADF and AEF share side AF, have AD = AE and DF = EF—three matching pairs of sides. SSS forces them to be congruent, so the angles at A must be equal. The bisector falls out automatically.
Conclusion: Here's why AF bisects the angle: We construct points D and E with AD = AE (by Prop 3). We construct equilateral triangle DEF on DE, so DF = EF (by Prop 1). Now compare triangles ADF and AEF. Side AD = side AE (by construction). Side DF = side EF (equilateral triangle DEF). Side AF = side AF (shared—it's the same line segment). By SSS (Proposition 8), △ADF ≅ △AEF. Corresponding angles of congruent triangles are equal, so ∠DAF = ∠EAF. Therefore line AF bisects ∠DAE, which is the same as ∠BAC. ✓
Key Moves
- Choose any point D on one side of the angle
- Mark point E on the other side so AD = AE (Proposition 3)—this is our first equal pair
- Construct equilateral triangle DEF on segment DE (Proposition 1)—this gives us DF = EF, our second equal pair
- Draw line AF—this is the shared side, our third equal pair (AF = AF)
- We now have three pairs: AD = AE, DF = EF, AF = AF
- By SSS (Proposition 8), △ADF ≅ △AEF
- Therefore ∠DAF = ∠EAF (corresponding angles of congruent triangles) ✓
Try It Yourself
Draw a wide angle (say about 120 degrees) and perform the construction: equal lengths on both sides, equilateral triangle, connect to the vertex. Now try it on a very narrow angle (about 20 degrees). Does the construction still work? What about a right angle—does your bisector look like 45 degrees?
Proof Challenge
Available Justifications
Given: ∠BAC. Take an arbitrary point D on AB.
Cut off AE from AC such that AE = AD
Join DE
Construct equilateral triangle DEF on DE
Join AF
In triangles ADF and AEF: AD = AE, DF = EF, AF = AF (common). Therefore △ADF ≅ △AEF by SSS.
Therefore ∠DAF = ∠EAF — the angle ∠BAC is bisected by AF.
Curriculum Materials
Get the Teaching Materials
Structured lesson plan, student worksheet, and answer key for Proposition 9.
- ✓Lesson Plan (prop-09-lesson-plan.pdf)
- ✓Student Worksheet (prop-09-worksheet.pdf)
- ✓Answer Key (prop-09-answer-key.pdf)
Why It Matters
Angle bisection is a fundamental construction. It lets us divide any angle in half using only compass and straightedge. This is essential for many later constructions.
Modern connection: Angle bisectors appear in optics (light reflection), in design (symmetric patterns), and in navigation. The angle bisector of a triangle has special properties—it divides the opposite side proportionally.
Historical note: The ancient Greeks were very interested in angle trisection (dividing into thirds), which turns out to be IMPOSSIBLE with compass and straightedge alone. Bisection is easy; trisection is provably impossible!
Discussion Questions
- Why does creating an equilateral triangle help us bisect the angle?
- Could you bisect an angle if you only had a straightedge (no compass)?
- Why is angle TRIsection (dividing into three) so much harder than bisection?
Euclid's Original Proof
Given: Angle BAC to be bisected. Construction: Take any point D on AB [Post. 1]. Cut off AE from AC equal to AD [I.3]. Join DE [Post. 1]. Construct equilateral triangle DEF on DE [I.1]. Join AF [Post. 1]. Claim: AF bisects angle BAC. Proof: In triangles ADF and AEF: - AD = AE (by construction) - DF = EF (sides of equilateral triangle) - AF is common Therefore △ADF ≅ △AEF by SSS [I.8]. Hence ∠DAF = ∠EAF. Therefore AF bisects angle BAC. Q.E.F.
What's Next
We can now bisect any angle. The natural companion question: can we bisect a line segment—find its exact midpoint? Proposition 10 shows how, and the trick is surprisingly elegant: build an equilateral triangle on the segment, then bisect the top angle. The bisector drops straight to the midpoint.