Proposition 7 of 48 • Theorem

Given two straight lines constructed from the ends of a straight line and meeting in a point, there cannot be constructed from the ends of the same straight line, and on the same side of it, two other straight lines meeting in another point and equal to the former two respectively.

If you have two lines meeting at a point above a base line, you can't draw two OTHER lines from the same base, on the same side, meeting at a DIFFERENT point, with the pairs having matching lengths.

Before You Read

Imagine you have a base line and you build a triangle on it with specific side lengths. Now try to build a DIFFERENT triangle on the same side of that base with the same side lengths but meeting at a different point. Can you? Grab two sticks of fixed lengths and a base—can those sticks meet at more than one point on the same side?

Browse All Foundations

All Foundations

Definitions (23)

Postulates (5)

Common Notions (5)

Browse All Propositions

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 base AB with lines AC and BC meeting at C. The claim: you cannot have another point D (on the same side) where AD = AC and BD = BC.

Approach: The strategy is another contradiction argument. Suppose such a rogue point D does exist. Connect C to D. Now look at what happens: triangle ACD is isosceles (AC = AD), so its base angles are equal by Proposition 5. Triangle BCD is also isosceles (BC = BD), so its base angles are equal too. But here's the catch—the position of D forces one of those angle pairs to be nested inside the other. You end up with an angle that must be simultaneously bigger and smaller than another angle. That's impossible, so D can't exist.

Conclusion: Assume D exists with AC = AD and BC = BD. Connect C to D. From isosceles triangle ACD: ∠ACD = ∠ADC (Prop 5). From isosceles triangle BCD: ∠BCD = ∠BDC (Prop 5). Euclid then argues from the diagram: the position of D forces ∠ACD > ∠BCD, and 'much more' ∠BDC > ∠BCD. But we just said ∠BDC = ∠BCD—contradiction. Therefore no such point D can exist. ✓

Key Moves

Assume (for contradiction) that point D exists on the same side of AB as C, with AC = AD and BC = BD
Join C to D (Postulate 1)
Since AC = AD, triangle ACD is isosceles, so ∠ACD = ∠ADC (Proposition 5)
Since ∠ACD = ∠ADC, we know ∠ACD is NOT less than ∠ADC
Since BC = BD, triangle BCD is isosceles, so ∠BCD = ∠BDC (Proposition 5)
But the position of D forces ∠ACD > ∠BCD (part vs. whole), so ∠ADC > ∠BDC
Yet also ∠BDC > ∠ADC (part vs. whole from D's side)—contradiction (Common Notion 5)
Therefore no such point D can exist on the same side of AB ✓

Try It Yourself

Take a ruler and mark a base segment. Now pick two lengths (say 5 cm and 7 cm). Using a compass, draw arcs from each end of the base with those radii. How many intersection points do you get on one side? Try different length combinations—do you ever get two intersection points on the same side?

Proof Challenge

Available Justifications

Given: On base AB, triangles ACB and ADB on the same side with AC = AD and BC = BD.

Drag justification

Join CD.

Drag justification

Since AC = AD, triangle ACD is isosceles

Drag justification

Therefore ∠ACD = ∠ADC (base angles of isosceles triangle ACD)

Drag justification

Since BC = BD, triangle BCD is isosceles, so ∠BCD = ∠BDC

Drag justification

But ∠ACD > ∠BCD while ∠ADC < ∠BDC (or vice versa), so equal angles cannot equal unequal angles — contradiction.

Drag justification

0 of 6 steps completed

Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 7.

✓Lesson Plan (prop-07-lesson-plan.pdf)
✓Student Worksheet (prop-07-worksheet.pdf)
✓Answer Key (prop-07-answer-key.pdf)

Browse Curriculum Bundles

Why It Matters

This is a uniqueness lemma—it says that given the right conditions, there's only ONE possible configuration. It's a technical result needed to prove Proposition 8 (SSS congruence).

Modern connection: This is related to the rigidity of triangles. Once you fix three side lengths, the triangle is completely determined (up to reflection). This is why triangles are used in structures—they don't flex.

Historical note: Euclid's proof here has been criticized for relying too heavily on the diagram. His 'much greater' step assumes angle relationships that he doesn't fully justify—you're supposed to 'see' them from the picture. A cleaner modern argument: if AC = AD and BC = BD, then D lies on both the circle centered at A (radius AC) and the circle centered at B (radius BC). These circles meet at exactly two points—C and its reflection across line AB. That reflection is on the opposite side of AB from C, so no such D exists on the same side.

Discussion Questions

Why does Euclid need this technical lemma? What's it setting up?
What if D were on the OPPOSITE side of AB from C? Would the proposition still hold?
How does this relate to the idea that triangles are 'rigid'?

Euclid's Original Proof

Given: On line AB, two distinct triangles ACB and ADB are constructed on the same side, where AC = AD and BC = BD.

Claim: This is impossible (C and D must be the same point).

Proof: Join CD [Post. 1].

Since AC = AD, triangle ACD is isosceles, so ∠ACD = ∠ADC [I.5].

Since BC = BD, triangle BCD is isosceles, so ∠BCD = ∠BDC [I.5].

But ∠ACD > ∠BCD (since D is between the rays). So ∠ADC > ∠BDC.

Yet also ∠BDC > ∠ADC (since C is between the rays)—contradiction.

Therefore two distinct points C and D with these properties cannot exist on the same side of AB.

Q.E.D.

What's Next

This uniqueness result is the missing piece for a big theorem. If three side lengths can only produce one triangle shape (on a given side), then matching all three sides should force two triangles to be identical. That's exactly what Proposition 8 proves—SSS congruence.