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.

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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: The proof copies the smaller angle inside the larger one. Copy ∠EDF at A onto ray AB (Proposition 23) — since ∠BAC > ∠EDF, the copied ray falls inside ∠BAC — and mark G on it with AG = DF (Proposition 3). Triangle ABG is then congruent to triangle DEF by SAS, so BG = EF, and the comparison happens entirely inside triangle ABC: the isosceles triangle ACG and two whole-greater-than-part observations show that in triangle BGC the angle at G exceeds the angle at C, so BC > BG = EF. (Euclid's own text runs the mirror-image construction on triangle DEF instead, building ∠EDG = ∠BAC there; the argument is the same.)

Conclusion: Copy ∠EDF at A onto ray AB (Proposition 23); since ∠BAC > ∠EDF, the copied ray falls inside ∠BAC. Mark G on it with AG = DF = AC (Proposition 3). Join BG and CG (Postulate 1). In triangles ABG and DEF: AB = DE, AG = DF, and ∠BAG = ∠EDF, so BG = EF (Proposition 4). Triangle ACG is isosceles (AG = AC), so ∠ACG = ∠AGC (Proposition 5). Now ray CB falls within ∠ACG, so ∠ACG > ∠BCG; and ray GA falls within ∠BGC, so ∠BGC > ∠AGC (Common Notion 5). Chaining: ∠BGC > ∠AGC = ∠ACG > ∠BCG. In triangle BGC the greater angle is subtended by the greater side, so BC > BG (Proposition 19). But BG = EF. Therefore BC > EF. ✓

Key Moves

Given AB = DE, AC = DF, and ∠BAC > ∠EDF
Copy ∠EDF at A onto ray AB (Proposition 23); since ∠BAC > ∠EDF, the copied ray falls inside ∠BAC
Mark G on the copied ray with AG = DF = AC (Proposition 3). Join BG and CG (Postulate 1).
In triangles ABG and DEF: AB = DE, AG = DF, ∠BAG = ∠EDF — so BG = EF (Proposition 4)
Triangle ACG is isosceles (AG = AC), so ∠ACG = ∠AGC (Proposition 5)
∠ACG > ∠BCG (ray CB falls within ∠ACG) and ∠BGC > ∠AGC (ray GA falls within ∠BGC), so ∠BGC > ∠BCG (Common Notion 5)
In triangle BGC, BC > BG (Proposition 19); and BG = EF, 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

Copy ∠EDF at vertex A onto ray AB. Since ∠BAC > ∠EDF, the copied ray falls inside ∠BAC.

Drag justification

On the copied ray, mark G with AG = DF (and DF = AC, so AG = AC).

Drag justification

Join BG and join CG.

Drag justification

In triangles ABG and DEF: AB = DE, AG = DF, and ∠BAG = ∠EDF (the copied angle). Therefore BG = EF.

Drag justification

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

Drag justification

∠ACG > ∠BCG, since ray CB falls within ∠ACG (the whole is greater than the part). And ∠ACG = ∠AGC, so ∠AGC > ∠BCG.

Drag justification

Also ∠BGC > ∠AGC, since ray GA falls within ∠BGC. Chaining: ∠BGC > ∠AGC = ∠ACG > ∠BCG, so ∠BGC > ∠BCG.

Drag justification

In triangle BGC, the greater angle ∠BGC is subtended by the greater side: BC > BG. And BG = EF, therefore BC > EF.

Drag justification

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Curriculum Materials

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The lesson plan, student worksheet, and answer key for Proposition 24 come with the curriculum bundles.

✓Included in Foundations (Propositions 1–26)
✓or the Complete Collection (all 48)

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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).

Going deeper

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.