Proposition 24 of 48 • Theorem

If two triangles have two sides equal to two sides respectively, but have one of the angles contained by the equal straight lines greater than the other, then they also have the base greater than the base.

If two triangles have two pairs of equal sides but one has a larger included angle than the other, then the triangle with the larger angle also has the longer base. (The 'hinge theorem': opening the hinge wider pushes the opposite side out further.)

Before You Read

Think of a door hinge with two arms of fixed length. If you open the hinge wider, the distance between the free ends grows longer. That seems obvious—but can you actually prove it? Proposition 24, the hinge theorem, says: if two triangles share two pairs of equal sides but one has a larger included angle, then it also has the longer opposite 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 two triangles ABC and DEF where AB = DE and AC = DF, but ∠BAC > ∠EDF. We want to prove that BC > EF.

Approach: Euclid places DF along AC (since AC = DF) and constructs the comparison geometrically. Since ∠BAC > ∠EDF, ray DE falls inside angle BAC. He constructs DG equal to DF such that ∠EDG = ∠BAC, positioning G so that triangle DEG corresponds to triangle ABC by SAS. Then he uses the isosceles triangle theorem and angle comparisons to show EF < EG = BC.

Conclusion: Place ∠EDF on ∠BAC so that DE lies along AB. Since DE = AB, the placed D coincides with A and placed E with B. Since ∠BAC > ∠EDF, the placed ray DF falls inside ∠BAC. Mark G on this ray with AG = DF = AC (Proposition 3). Join BG and CG (Postulate 1). Triangle ACG is isosceles (AG = AC), so ∠ACG = ∠AGC (Proposition 5). Since G is inside ∠BAC, ray CG lies inside ∠BCA, so ∠BCA > ∠ACG = ∠AGC (Common Notion 5). In triangle BGC, ∠BCA > ∠BGC, so BC > BG (Proposition 19). But BG = EF (the base of the placed copy of DEF). Therefore BC > EF. ✓

Key Moves

Given AB = DE, AC = DF, and ∠BAC > ∠EDF
Place ∠EDF on ∠BAC so DE coincides with AB (Proposition 23). Placed ray DF falls inside ∠BAC.
Mark G on the interior ray with AG = DF = AC (Proposition 3). Join BG and CG (Postulate 1).
Triangle ACG is isosceles (AG = AC), so ∠ACG = ∠AGC (Proposition 5)
∠BCA > ∠ACG since ray CG lies inside ∠BCA, so ∠BCA > ∠AGC (Common Notion 5)
In triangle BGC, ∠BCA > ∠BGC, so BC > BG (Proposition 19)
BG = EF (placed copy), therefore BC > EF ✓

Try It Yourself

Draw two triangles ABC and DEF where AB = DE = 6 cm and AC = DF = 8 cm, but make angle BAC noticeably larger than angle EDF. Now measure BC and EF: which is longer? Try a few different angle sizes and confirm the pattern. Then try to imagine a case where the wider angle produces a shorter base—you will find you cannot.

Proof Challenge

Available Justifications

Place ∠EDF on ∠BAC so that DE lies along AB. Since DE = AB, placed D coincides with A and placed E with B.

Drag justification

Since ∠BAC > ∠EDF, the placed ray DF falls inside ∠BAC. On this ray, mark G with AG = DF = AC.

Drag justification

Join BG and join CG.

Drag justification

Triangle ACG is isosceles (AG = AC), so ∠ACG = ∠AGC.

Drag justification

∠BCA > ∠ACG (since G is inside ∠BAC, ray CG lies inside ∠BCA), so ∠BCA > ∠AGC.

Drag justification

In triangle BGC, ∠BCA > ∠BGC, so BC > BG (Prop 19). But BG = EF (placed copy). Therefore BC > EF.

Drag justification

0 of 6 steps completed

Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 24.

✓Lesson Plan (prop-24-lesson-plan.pdf)
✓Student Worksheet (prop-24-worksheet.pdf)
✓Answer Key (prop-24-answer-key.pdf)

Browse Curriculum Bundles

Why It Matters

The hinge theorem captures an essential geometric intuition: widening the angle between two fixed-length sides pushes the opposite side outward, making it longer. This is one of the key comparison theorems for triangles and directly underpins its converse (Prop 25).

Modern connection: The hinge theorem models real-world mechanisms: a door hinge, a pair of scissors, a robot arm. In mechanical engineering, knowing that a wider opening angle means a greater reach is fundamental to linkage design. In navigation, it explains why changing your bearing slightly can dramatically change how far you end up from your starting point.

Historical note: This proposition is sometimes called the 'open mouth theorem' or 'scissor theorem' because of the vivid hinge analogy. Euclid's proof is considered one of the more intricate in Book I, requiring careful placement of the triangles and subtle angle comparisons.

Discussion Questions

Why is this called the 'hinge theorem'? Can you think of a physical object that demonstrates this principle?
The proof requires careful placement of one triangle relative to the other. Why can't we just use algebra or the law of cosines to prove this directly?
Propositions 24 and 25 are converses. Together, what complete relationship do they establish between the included angle and the opposite side?

Euclid's Original Proof

If two triangles have the two sides equal to two sides respectively, but have the one of the angles contained by the equal straight lines greater than the other, they will also have the base greater than the base.

Let ABC, DEF be two triangles having the two sides AB, AC equal to the two sides DE, DF respectively, namely AB to DE, and AC to DF, and let the angle at A be greater than the angle at D;

I say that the base BC is also greater than the base EF.

For, since the angle BAC is greater than the angle EDF, let there be constructed, on the straight line DE, and at the point D on it, the angle EDG equal to the angle BAC; [I.23]

let DG be made equal to either of the two straight lines AC, DF, [I.3]

and let EG, FG be joined. [Post. 1]

Then, since AB is equal to DE, and AC to DG, the two sides BA, AC are equal to the two sides ED, DG, respectively;

and the angle BAC is equal to the angle EDG;

therefore the base BC is equal to the base EG. [I.4]

Again, since DF is equal to DG, the angle DGF is also equal to the angle DFG; [I.5]

therefore the angle DFG is greater than the angle EGF.

Therefore the angle EFG is much greater than the angle EGF. [C.N. 5]

And, since EFG is a triangle having the angle EFG greater than the angle EGF, and the greater angle is subtended by the greater side, [I.19]

the side EG is also greater than EF.

But EG is equal to BC.

Therefore BC is also greater than EF.

Therefore etc.

Q.E.D.

What's Next

The hinge theorem establishes that a larger included angle means a longer opposite side. But does it go the other way? If one base is longer than the other, must its included angle be larger? That converse—proved by eliminating the other two logical possibilities—is exactly what Proposition 25 delivers.