Proposition 12 of 48 • Construction

To draw a straight line perpendicular to a given infinite straight line from a given point not on it.

Given a line and a point not on it, drop a perpendicular from the point to the line.

Before You Read

Now the challenge flips: you have a point hovering above a line, and you need to drop a perpendicular straight down to the line—hitting it at a perfect right angle. Your point isn't on the line, so Proposition 11's trick won't work directly. How would you find that foot of the perpendicular?

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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 an infinite line AB and a point C that is not on it. We need to draw a line from C straight down to AB, hitting it at a right angle.

Approach: The strategy: draw a circle centered at C large enough to cross line AB in two places (G and E). Then find the midpoint H of GE. The line CH will be perpendicular to AB, because H is equidistant from G and E, and C is equidistant from G and E (both are radii).

Conclusion: Here's why CH is perpendicular to AB: The circle centered at C crosses AB at G and E, so CG = CE (both are radii, Def 15). We bisect GE at H (Prop 10), so GH = HE. Now compare triangles CHG and CHE. Side CG = side CE (radii). Side GH = side HE (H is midpoint). Side CH = side CH (shared). By SSS (Proposition 8), △CHG ≅ △CHE. Therefore ∠CHG = ∠CHE. These are adjacent angles on line AB, and they're equal—so by Definition 10, they're right angles. CH is perpendicular to AB. ✓

Key Moves

Take point D on the opposite side of AB from C
Draw a circle centered at C with radius CD (Postulate 3)—it crosses AB at G and E
CG = CE because both are radii of the same circle (Definition 15)
Bisect GE at point H (Proposition 10), so GH = HE
Draw CH (Postulate 1)
In triangles CHG and CHE: CG = CE, GH = HE, CH = CH
By SSS (Proposition 8), △CHG ≅ △CHE
Therefore ∠CHG = ∠CHE—equal adjacent angles are right angles (Definition 10) ✓

Try It Yourself

Draw a line and a point clearly above it. Use a compass to draw a circle centered at your point that cuts the line in two places, then find the midpoint of that chord—that's the foot of the perpendicular. Check your result with a protractor. Try placing your external point at different heights and distances.

Proof Challenge

Available Justifications

Take point D on the opposite side of line AB from C

Drag justification

Draw circle with center C and radius CD, cutting AB at G and E

Drag justification

Bisect segment GE at point H

Drag justification

Draw lines CG, CH, and CE

Drag justification

CG = CE (radii of same circle), GH = EH (H bisects GE), CH = CH

Drag justification

∠CHG = ∠CHE and they are adjacent on line GE, so CH ⊥ AB

Drag justification

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Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 12.

✓Lesson Plan (prop-12-lesson-plan.pdf)
✓Student Worksheet (prop-12-worksheet.pdf)
✓Answer Key (prop-12-answer-key.pdf)

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Why It Matters

Dropping a perpendicular from a point to a line is essential for measuring distances. The shortest distance from a point to a line IS the perpendicular distance. This construction appears constantly in later geometry.

Modern connection: Every time you measure the height of a building, the depth of a well, or the altitude of a triangle, you're using a perpendicular from a point to a line. GPS altitude measurements, architectural blueprints, and physics force decompositions all rely on this concept.

Historical note: Notice the difference from Proposition 11: there the point was ON the line, here it's OFF the line. Euclid treats these as separate problems because the constructions are quite different. Prop 11 uses an equilateral triangle; Prop 12 uses a circle and a midpoint.

Discussion Questions

Why does Euclid need the point D on the opposite side? What role does it play?
How do you know the circle will actually cross the line in two places?
What's the difference between this construction and Proposition 11? When would you use each?

Euclid's Original Proof

Let AB be the given infinite straight line, and C the given point which is not on it.

Thus it is required to draw from the point C a straight line perpendicular to the straight line AB.

For let a point D be taken at random on the other side of the straight line AB, and with centre C and distance CD let the circle EGF be described; [Post. 3]
let the straight line EG be bisected at H, [I.10]
and let the straight lines CG, CH, CE be joined. [Post. 1]

I say that CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it.

For, since GH is equal to HE, and HC is common,
the two sides GH, HC are equal to the two sides EH, HC respectively;
and the base CG is equal to the base CE; [Def. 15]
therefore the angle CHG is equal to the angle CHE. [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, and the straight line standing on the other is called a perpendicular to that on which it stands. [Def. 10]

Therefore CH has been drawn perpendicular to the given infinite straight line AB from the given point C which is not on it.

(Being) what it was required to do.

What's Next

We can now erect perpendiculars from points on a line and drop them from points off a line. Proposition 13 shifts from construction to theorem: whenever any line meets another, what can we say about the two angles it creates? The answer—they always sum to two right angles—is one of geometry's most used facts.