Proposition 10 of 48 • Construction

To bisect a given finite straight line.

Given any line segment, find its exact midpoint.

Before You Read

Draw a line segment on your paper. Now find its exact midpoint—but you can't use a ruler to measure. You only have a compass and straightedge. Hint: you already know how to build equilateral triangles and bisect angles. Can you combine those to split a line in half?

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All Foundations

Definitions (23)

Postulates (5)

Common Notions (5)

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All Propositions

Basic Constructions (5)

Triangle Fundamentals (5)

Perpendiculars & Angles (5)

Exterior Angles & Inequalities (5)

Interior Triangles & Angle Copying (5)

Parallel Lines (5)

Parallel Constructions & Parallelograms (5)

Area Theorems (5)

Area Applications (4)

Grand Finale (4)

What Euclid Is Doing

Setup: We have line segment AB and want to find its exact middle point.

Approach: The strategy is to turn a line-bisection problem into an angle-bisection problem—which we already know how to solve. Build an equilateral triangle on the segment (Prop 1), then bisect the top angle (Prop 9). The bisector drops straight down to the midpoint. Why? Because the equilateral triangle is perfectly symmetric, and the angle bisector is the axis of symmetry. It has to hit the base at the halfway mark.

Conclusion: The bisector hits AB at point D. Since triangle ACB is equilateral, CA = CB. The angle bisector creates two triangles ACD and BCD with CA = CB, ∠ACD = ∠BCD, and CD shared. By SAS, they're congruent, so AD = BD.

Key Moves

Construct equilateral triangle ABC on segment AB (Proposition 1). Now CA = CB = AB (Definition 20).
Bisect angle ACB: construct line CD that splits ∠ACB into two equal parts (Proposition 9)
Mark point D where the bisector CD meets line AB (Postulate 1)
In triangles ACD and BCD: CA = CB (sides of equilateral triangle, Definition 20)
∠ACD = ∠BCD (CD bisects angle ACB, by construction via Proposition 9)
CD = CD (common side—shared by both triangles)
By SAS (Proposition 4): △ACD ≅ △BCD, so all corresponding parts are equal
Therefore AD = BD (corresponding sides of congruent triangles), and D is the midpoint ✓

Try It Yourself

Draw a line segment of any length and perform the full construction: equilateral triangle, angle bisection, find the midpoint. Now measure both halves with a ruler to check. Try it on a very short segment and a very long one. Does the construction always land on the exact middle?

Proof Challenge

Available Justifications

Given: Line segment AB. Construct equilateral triangle ABC on AB.

Drag justification

Bisect ∠ACB with line CD, where D is on AB

Drag justification

In triangles ACD and BCD: CA = CB (equilateral triangle), ∠ACD = ∠BCD (bisected angle), CD = CD (common)

Drag justification

Therefore △ACD ≅ △BCD by SAS

Drag justification

Therefore AD = BD — point D is the midpoint of AB.

Drag justification

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Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 10.

✓Lesson Plan (prop-10-lesson-plan.pdf)
✓Student Worksheet (prop-10-worksheet.pdf)
✓Answer Key (prop-10-answer-key.pdf)

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Why It Matters

Finding midpoints is essential for countless constructions. You need midpoints to draw perpendicular bisectors, to find centers, to divide things evenly. This is one of the most-used constructions.

Modern connection: Midpoints are fundamental in computer graphics (averaging coordinates), in physics (center of mass calculations), and in statistics (finding medians). The concept of 'halfway' pervades mathematics.

Historical note: Notice how the propositions build on each other: Prop 10 uses Prop 9 (angle bisection), which uses Prop 8 (SSS), which uses Prop 7, which uses Prop 5, which uses Prop 4. This logical chain is the beauty of Euclid's approach.

Discussion Questions

Why do we need an equilateral triangle? Would any isosceles triangle work?
Could you find the midpoint using only circles (no angle bisection)?
What's the connection between bisecting an angle and bisecting a line?

Euclid's Original Proof

Given: Finite straight line AB to be bisected.

Construction: Construct equilateral triangle ABC on AB [I.1]. Bisect angle ACB by line CD meeting AB at D [I.9].

Claim: D is the midpoint of AB.

Proof: In triangles ACD and BCD:
- AC = BC (triangle ABC is equilateral)
- ∠ACD = ∠BCD (CD bisects angle ACB)
- CD = CD (common side)

Therefore △ACD ≅ △BCD by SAS [I.4].

Hence AD = BD.

Therefore the straight line AB has been bisected at D.

Q.E.F.

What's Next

We can now bisect both angles and line segments. Proposition 11 tackles the next fundamental construction: given a point sitting on a line, how do you draw a perpendicular through that point? It builds directly on the midpoint-finding technique from this proposition.