Proposition 11 of 48 • Construction

To draw a straight line at right angles to a given straight line from a given point on it.

Given a point on a line, construct a perpendicular line through that point.

Before You Read

You're standing on a straight road and need to build a wall that meets the road at a perfect right angle—not approximately, but exactly 90 degrees. You have only a compass and straightedge. Can you guarantee perfection without any measuring device?

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Definitions (23)

Postulates (5)

Common Notions (5)

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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 a straight line AB and a point C on it. We need to draw a line through C that is perpendicular to AB—that is, making right angles on both sides.

Approach: The idea is to create a symmetric situation around point C. We mark equal distances on both sides of C (giving us D and E with CD = CE), then build an equilateral triangle on DE. The apex of that triangle, point F, must be directly above C by symmetry.

Conclusion: Here's the proof that CF is perpendicular to AB: We chose D and E so that CD = CE (Prop 3). We built equilateral triangle DEF on DE (Prop 1), so DF = EF. Now compare triangles DCF and ECF. Side DC = side EC (by construction). Side DF = side EF (equilateral triangle). Side CF = side CF (shared). By SSS (Proposition 8), △DCF ≅ △ECF. Therefore ∠DCF = ∠ECF. But ∠DCF and ∠ECF are adjacent angles on line AB, and they're equal. By Definition 10, when a straight line standing on another makes equal adjacent angles, those angles are right angles. So CF is perpendicular to AB at C. ✓

Key Moves

Take any point D on AC
Cut off CE equal to CD on the other side (Proposition 3)—now C is the midpoint of DE
Construct equilateral triangle DEF on segment DE (Proposition 1)
Draw line CF (Postulate 1)
In triangles DCF and ECF: DC = EC, DF = EF, CF = CF
By SSS (Proposition 8), △DCF ≅ △ECF
Therefore ∠DCF = ∠ECF
Equal adjacent angles on a straight line are right angles (Definition 10) ✓

Try It Yourself

Draw a horizontal line and pick a point somewhere in the middle of it. Now perform Euclid's construction: mark two equal lengths on either side of your point, build an equilateral triangle on them, and connect its apex back to your point. Use a protractor to check—is that really 90 degrees?

Proof Challenge

Available Justifications

Take point D on ray CA; cut off CE = CD on ray CB

Drag justification

Construct equilateral triangle DEF on segment DE

Drag justification

Draw line CF

Drag justification

In triangles DCF and ECF: DC = EC, DF = EF, CF = CF — so △DCF ≅ △ECF

Drag justification

∠DCF = ∠ECF and they are adjacent on line DE, so each is a right angle

Drag justification

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Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 11.

✓Lesson Plan (prop-11-lesson-plan.pdf)
✓Student Worksheet (prop-11-worksheet.pdf)
✓Answer Key (prop-11-answer-key.pdf)

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Why It Matters

Drawing a perpendicular from a point on a line is one of the most fundamental constructions in geometry. It's the basis for coordinate systems, right angles in architecture, and any situation where you need to go 'straight up' from a surface.

Modern connection: Every building, bridge, and structure depends on perpendicularity. The walls of a house are perpendicular to the floor. Computer screens have perpendicular axes (x and y). The entire Cartesian coordinate system is built on perpendicular lines.

Historical note: This construction is beautifully simple: equal distances on each side, an equilateral triangle on top, and symmetry does the rest. The ancient Egyptians used a similar technique with knotted ropes to create right angles for building the pyramids.

Discussion Questions

Why is it important that D and E are equidistant from C? What happens if they aren't?
Could you construct a perpendicular without using an equilateral triangle? What other approach might work?
How does this construction relate to Proposition 10 (bisecting a line segment)?

Euclid's Original Proof

Let AB be the given straight line, and let C be a given point on it.

Thus it is required to draw from the point C a straight line at right angles to the straight line AB.

Let a point D be taken at random on AC; let CE be made equal to CD; [I.3]
on DE let the equilateral triangle FDE be constructed, [I.1]
and let FC be joined; [Post. 1]

I say that the straight line FC has been drawn at right angles to the given straight line AB from C the given point on it.

For, since DC is equal to CE, and CF is common,
the two sides DC, CF are equal to the two sides EC, CF respectively;
and the base DF is equal to the base EF;
therefore the angle DCF is equal to the angle ECF; [I.8]
and they are adjacent angles.

But, when a straight line set up on a straight line makes the adjacent angles equal to one another, each of the equal angles is right, [Def. 10]
and the straight line standing on the other is called a perpendicular to that on which it stands.

Therefore CF has been drawn at right angles to the given straight line AB from the given point C on it.

(Being) what it was required to do.

What's Next

Proposition 11 raises a perpendicular from a point on the line. But what if your point is not on the line at all—it's floating above it? Dropping a perpendicular from an external point turns out to need a different, equally elegant construction, and that's exactly what Proposition 12 delivers.