Euclid's WorkshopBook I
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Proposition 14 of 48 • Theorem

If with any straight line, and at a point on it, two straight lines not lying on the same side make the sum of the adjacent angles equal to two right angles, then the two straight lines are in a straight line with one another.

If two lines on opposite sides of a third line make angles that sum to 180 degrees at the meeting point, then those two lines form a single straight line.

Before You Read

Two line segments meet a straight line from opposite sides at the same point. Someone tells you the two angles they make with the line add up to exactly 180 degrees. Does that guarantee the two segments are really one continuous straight line, or could they still be bent slightly apart?

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A B C D E

What Euclid Is Doing

Setup: Line AB stands on a point B. On one side we have line BC, on the other side line BD, and ∠ABC + ∠ABD = two right angles. We want to prove that C, B, and D are collinear—that BC and BD form one straight line.

Approach: Euclid uses proof by contradiction. He assumes BD is NOT the straight-line extension of BC, meaning there's some other line BE that is the true extension. Then he shows this leads to a logical impossibility.

Conclusion: Suppose BD is NOT in line with BC. Then let BE be the extension of CB through B. By Proposition 13, ∠ABC + ∠ABE = two right angles (because CBE is a straight line). But we're given that ∠ABC + ∠ABD = two right angles. Subtracting, ∠ABE = ∠ABD (Common Notion 3: equals subtracted from equals give equals). But ∠ABD is clearly greater than ∠ABE since ∠ABD contains ∠ABE plus extra (Common Notion 5: the whole is greater than the part). So we have ∠ABE both equal to and less than ∠ABD—a contradiction. Therefore our assumption was wrong, and BD really is the extension of BC. ✓

Key Moves

  1. Given: ∠ABC + ∠ABD = two right angles, with BC and BD on opposite sides of AB
  2. Assume for contradiction: BD is not in line with BC
  3. Let BE be the actual straight-line extension of CB through B
  4. By Proposition 13: ∠ABC + ∠ABE = two right angles (CBE is a straight line)
  5. But ∠ABC + ∠ABD = two right angles (given)
  6. Therefore ∠ABE = ∠ABD (Common Notion 3: subtract ∠ABC from both)
  7. But ∠ABD > ∠ABE since D is outside ∠ABE (Common Notion 5: whole > part)
  8. Contradiction: ∠ABE cannot both equal and be less than ∠ABD
  9. Therefore BD must be in line with BC ✓

Try It Yourself

Draw a horizontal line and a point on it. Draw one ray upward-left making a 130-degree angle with the line on the left side. Now draw another ray upward-right—what angle must it make to force the two rays into a straight line? Try drawing it 'almost' straight and measure: does your sum still hit exactly 180 degrees?

Proof Challenge

Available Justifications

1.

Assume ∠ABC + ∠ABE = two right angles

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2.

Suppose BE is not on the same line as BC — let BD be the extension of CB

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3.

Then ∠ABC + ∠ABD = two right angles

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4.

So ∠ABE = ∠ABD

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5.

But ∠ABD > ∠ABE since D ≠ E — contradiction, so BE and BC are collinear

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Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 14.

  • ✓Lesson Plan (prop-14-lesson-plan.pdf)
  • ✓Student Worksheet (prop-14-worksheet.pdf)
  • ✓Answer Key (prop-14-answer-key.pdf)
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Why It Matters

This is the converse of Proposition 13. Together they give a complete characterization: two angles on a line sum to 180° if and only if the outer rays form a straight line. This 'if and only if' relationship is a powerful tool for proving lines are collinear.

Modern connection: Collinearity testing is essential in computer graphics, robotics, and surveying. When you need to check if three points lie on the same line, one approach is to verify that the angles sum to 180°—exactly this proposition.

Historical note: This is Euclid's first proof by contradiction in the Elements. He assumes the opposite of what he wants to prove, derives an impossibility, and concludes the original statement must be true. This technique (reductio ad absurdum) becomes a staple of his method.

Discussion Questions

  • Why does Euclid use proof by contradiction here instead of a direct proof?
  • How does this proposition relate to Proposition 13? Why do we need both?
  • Can you think of a situation where you'd need to prove three points are on the same line?
Euclid's Original Proof
If with any straight line, and at a point on it, two straight lines not lying on the same side make the adjacent angles equal to two right angles, the two straight lines will be in a straight line with one another.

For with the straight line AB, and at the point B on it, let the two straight lines BC, BD not lying on the same side make the adjacent angles ABC, ABD equal to two right angles;

I say that BD is in a straight line with CB.

For, if BD is not in a straight line with BC, let BE be in a straight line with CB.

Then, since the straight line AB stands on the straight line CBE,
the angles ABC, ABE are equal to two right angles. [I.13]

But the angles ABC, ABD are also equal to two right angles; [given]
therefore the angles CBA, ABE are equal to the angles CBA, ABD. [Post. 4 and C.N. 1]

Let the angle CBA be subtracted from each;
therefore the remaining angle ABE is equal to the remaining angle ABD, [C.N. 3]
the less to the greater: which is impossible.

Therefore BE is not in a straight line with CB.

Similarly we can prove that neither is any other straight line except BD.
Therefore BD is in a straight line with CB.

Q.E.D.

What's Next

Proposition 14 gives us a way to prove that three points are collinear by checking angle sums. Right on its heels comes Proposition 15—vertical angles—which uses the same angle-sum machinery to prove that when two lines cross, the opposite angles are always equal, a result you've probably assumed your whole life.