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Proposition 48 of 48 • Theorem

If in a triangle the square on one of the sides equals the sum of the squares on the remaining two sides of the triangle, then the angle contained by the remaining two sides of the triangle is right.

The converse of the Pythagorean theorem: if a triangle has the property that the square on one side equals the sum of the squares on the other two sides, then the angle between those two sides must be a right angle. Together with Proposition 47, this gives a complete 'if and only if' characterization: a triangle is right-angled exactly when the Pythagorean relation holds among its sides.

Before You Read

Proposition 47 told you that a right angle forces a² + b² = c² among the sides. Now flip it: if some triangle happens to satisfy a² + b² = c², does that guarantee the angle between sides a and b is exactly 90°? Could a triangle accidentally satisfy the equation without having a right angle?

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90°? constructed reference A B C D c b a Given: a² = b² + c² → ∠A = 90°

What Euclid Is Doing

Setup: In triangle ABC, suppose the square on side BC equals the sum of the squares on sides BA and AC. That is, BC² = BA² + AC². We must prove that angle BAC is a right angle.

Approach: Euclid uses a comparison strategy. He constructs a brand-new right triangle with the same two legs as triangle ABC, then applies the Pythagorean theorem (Proposition 47) to the new triangle to compute its hypotenuse. The hypothesis tells us the hypotenuses of both triangles are equal, so by SSS congruence (Proposition 8) the triangles are congruent and the angle in question must be right.

Conclusion: Draw line AD perpendicular to AC at A, and make AD equal to AB (Proposition 3). Join DC. Since angle DAC is right (by construction), triangle DAC is a right triangle. By Proposition 47 (the Pythagorean theorem), DC² = DA² + AC². But DA = AB, so DC² = AB² + AC². By our hypothesis, BC² = AB² + AC². Therefore DC² = BC², hence DC = BC. Now in triangles DAC and BAC: DA = BA (by construction), AC is common, and DC = BC (just proved). By Proposition 8 (SSS), triangle DAC is congruent to triangle BAC. Therefore angle DAC = angle BAC. But angle DAC is a right angle (by construction). Therefore angle BAC is a right angle. ✓

Key Moves

  1. Given: triangle ABC where BC² = BA² + AC².
  2. At A, draw AD perpendicular to AC (Proposition 11).
  3. Cut off AD = AB (Proposition 3).
  4. Join DC.
  5. Triangle DAC has a right angle at A, so by Proposition 47: DC² = DA² + AC².
  6. Since DA = AB: DC² = AB² + AC².
  7. By hypothesis: BC² = AB² + AC².
  8. Therefore DC² = BC², so DC = BC.
  9. In triangles DAC and BAC: DA = BA, AC = AC, DC = BC.
  10. By SSS congruence (Proposition 8): triangle DAC is congruent to triangle BAC.
  11. Therefore angle BAC = angle DAC = 90° ✓

Try It Yourself

Check several triangles: measure their sides with a ruler, compute whether a² + b² = c² holds, and then measure the angle between the two shorter sides with a protractor. Does every triangle satisfying the equation turn out to have a 90° angle? Now try the classic 3-4-5 triangle—verify both that 9 + 16 = 25 and that the enclosed angle is a right angle.

Proof Challenge

Available Justifications

1.

Given: Triangle ABC where BC² = AB² + AC². From point A, draw AD at right angles to AC, and make AD equal to AB.

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2.

Cut off AD equal to AB on the perpendicular.

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3.

Join DC. Since angle DAC is a right angle, apply Prop 47: DC² = DA² + AC².

Drag justification
4.

Since DA = AB, we have DC² = AB² + AC². But we are given BC² = AB² + AC². Therefore DC² = BC², so DC = BC.

Drag justification
5.

In triangles DAC and BAC: DA = BA, AC is common, DC = BC. Therefore triangle DAC is congruent to triangle BAC.

Drag justification
6.

Therefore angle DAC = angle BAC. But angle DAC is a right angle. Therefore angle BAC is a right angle.

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Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 48.

  • ✓Lesson Plan (prop-48-lesson-plan.pdf)
  • ✓Student Worksheet (prop-48-worksheet.pdf)
  • ✓Answer Key (prop-48-answer-key.pdf)
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Why It Matters

Proposition 48 completes the logical circle opened by Proposition 47: a triangle is right-angled if and only if the square on one side equals the sum of the squares on the other two. This 'if and only if' characterization — the Pythagorean theorem and its converse together — is the definitive statement about right triangles and the perfect conclusion to Book I. It is also Euclid's demonstration that the relationship between right angles and the sum-of-squares property goes both ways: right angles produce the property, and the property guarantees right angles.

Modern connection: The converse of the Pythagorean theorem is used constantly in practice: when you want to verify that an angle is exactly 90 degrees, you check the 3-4-5 relationship (or any Pythagorean triple). Builders and surveyors have used this technique for millennia — stretch a rope into a 3-4-5 triangle and the corner is guaranteed to be square. In abstract mathematics, the converse is what justifies defining orthogonality via the inner product: two vectors are orthogonal if and only if ||u + v||² = ||u||² + ||v||².

Historical note: The fact that Euclid ends Book I with the converse — not with the Pythagorean theorem itself — is a deliberate structural choice. It signals that mathematics is about complete characterization, not just one-directional results. The Pythagorean theorem tells you what happens when you have a right angle; the converse tells you how to recognize one. Together, they close Book I with a satisfying logical completeness that mirrors the axiomatic rigor with which it opened.

Discussion Questions

  • The proof constructs an auxiliary right triangle and then compares it to the given triangle using SSS (Proposition 8). Why not try to prove the angle is right directly, without introducing a second triangle?
  • Proposition 47 and 48 together give an 'if and only if.' In modern logic, this means the Pythagorean property is a necessary and sufficient condition for a right angle. Why does Euclid prove the two directions as separate propositions rather than combining them?
  • Builders verify right angles using the 3-4-5 triangle: if the sides are 3, 4, and 5 units, the angle opposite the longest side is right. This is a direct application of Proposition 48. What other Pythagorean triples could they use?
  • Book I ends here. Looking back, how do the 48 propositions fit together? Is the structure a single chain leading to Propositions 47-48, or more like a network with multiple threads?
Euclid's Original Proof
If in a triangle the square on one of the sides be equal to the squares on the remaining two sides of the triangle, the angle contained by the remaining two sides of the triangle is right.

For in the triangle ABC let the square on one side BC be equal to the squares on the sides BA, AC;

I say that the angle BAC is right.

For let AD be drawn from the point A at right angles to the straight line AC, [I.11]

let AD be made equal to BA, [I.3]

and let DC be joined.

Since DA is equal to AB, the square on DA is also equal to the square on AB.

Let the square on AC be added to each;

therefore the squares on DA, AC are equal to the squares on BA, AC.

But the square on DC is equal to the squares on DA, AC, for the angle DAC is right; [I.47]

and the square on BC is equal to the squares on BA, AC, for this is the hypothesis;

therefore the square on DC is equal to the square on BC,

so that the side DC is also equal to BC.

And, since DA is equal to AB, and AC is common, the two sides DA, AC are equal to the two sides BA, AC;

and the base DC is equal to the base BC;

therefore the angle DAC is equal to the angle BAC. [I.8]

But the angle DAC is right;

therefore the angle BAC is also right.

Therefore etc.

Q.E.D.