Proposition 34 of 48 • Theorem

In parallelogrammic areas the opposite sides and angles equal one another, and the diameter bisects the areas.

In any parallelogram, opposite sides are equal, opposite angles are equal, and the diagonal cuts it into two equal triangles. This proposition fully characterizes the basic properties of parallelograms.

Before You Read

A parallelogram looks symmetric—opposite sides seem equal, opposite corners seem to match—but looks can deceive in geometry. Can you actually prove all three of these properties from nothing but the fact that two pairs of sides are parallel?

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What Euclid Is Doing

Setup: ABCD is a parallelogram (a 'parallelogrammic area' — a quadrilateral with both pairs of opposite sides parallel: AB ∥ DC and AD ∥ BC). We must prove three things: (1) opposite sides are equal (AB = DC and AD = BC), (2) opposite angles are equal (∠DAB = ∠BCD and ∠ABC = ∠CDA), and (3) the diagonal BD bisects the area (triangle ABD = triangle CDB in area).

Approach: Euclid draws the diagonal BD (connecting opposite vertices A's neighbor B to D). Since AB is parallel to DC and AD is parallel to BC, the diagonal acts as a transversal for both pairs of parallels, generating two pairs of equal alternate interior angles. This gives enough angle equalities to prove the two triangles formed by the diagonal are congruent (ASA via Proposition 26), from which all three results follow.

Conclusion: Draw diagonal BD. Since AB is parallel to DC, the transversal BD creates equal alternate angles ∠ABD = ∠BDC (Proposition 29). Since AD is parallel to BC, the transversal BD creates equal alternate angles ∠ADB = ∠DBC (Proposition 29). The side BD is common to both triangles ABD and CDB. By ASA (Proposition 26): triangle ABD is congruent to triangle CDB. Therefore AB = DC and AD = BC (corresponding sides). Also ∠DAB = ∠BCD (corresponding angles), and since ∠ABD + ∠DBC = ∠ABC and ∠ADB + ∠BDC = ∠ADC, the opposite angles ∠ABC = ∠ADC follow from adding the equal parts (Common Notion 2). Finally, since the triangles are congruent, they are equal in area — the diagonal bisects the parallelogram. ✓

Key Moves

Draw diagonal BD in parallelogram ABCD (AB ∥ DC, AD ∥ BC), creating triangles ABD and CDB.
AB parallel to DC with transversal BD: alternate angles ∠ABD = ∠BDC (Proposition 29).
AD parallel to BC with transversal BD: alternate angles ∠ADB = ∠DBC (Proposition 29).
Side BD is common to both triangles.
By ASA (Proposition 26): triangle ABD is congruent to triangle CDB.
Corresponding sides: AB = DC and AD = BC (opposite sides equal) ✓
Corresponding angles give opposite angles equal (with Common Notion 2 for sums) ✓
Congruent triangles have equal area: diagonal BD bisects the parallelogram ✓

Try It Yourself

Draw a parallelogram of any shape—try a very squashed one—then draw one diagonal. Measure the two triangles you've created: their sides, their angles, their areas. Do both diagonals bisect the area equally? What happens when you draw the second diagonal instead?

Proof Challenge

Available Justifications

Given: Parallelogram ABCD (AB ∥ DC, AD ∥ BC). Draw diagonal BD.

Drag justification

∠ABD = ∠BDC (alternate angles, AB ∥ DC with transversal BD)

Drag justification

∠ADB = ∠DBC (alternate angles, AD ∥ BC with transversal BD)

Drag justification

In triangles ABD and CDB: ∠ABD = ∠BDC, BD = BD (common), ∠ADB = ∠DBC. So △ABD ≅ △CDB by ASA.

Drag justification

Therefore AB = DC and AD = BC (opposite sides are equal)

Drag justification

∠DAB = ∠BCD and ∠ABC = ∠CDA (opposite angles are equal)

Drag justification

△ABD = △CDB in area, so diagonal BD bisects the parallelogram

Drag justification

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Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 34.

✓Lesson Plan (prop-34-lesson-plan.pdf)
✓Student Worksheet (prop-34-worksheet.pdf)
✓Answer Key (prop-34-answer-key.pdf)

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Why It Matters

This is the fundamental theorem about parallelograms. Every subsequent area result in Book I — from Proposition 35 through to the Pythagorean theorem (Prop 47) — depends on these properties. The fact that the diagonal bisects the area is especially important: it is the key tool for comparing areas of parallelograms and triangles.

Modern connection: Parallelogram properties are foundational in physics and engineering. The parallelogram law of force addition (used to combine vectors) depends on opposite sides being equal and parallel. In structural engineering, parallelogram linkages maintain these properties under motion, which is why they appear in everything from scissor lifts to pantographs.

Historical note: Euclid uses the term 'parallelogrammic area' rather than just 'parallelogram,' emphasizing that these are regions with area, not just outlines. This distinction becomes critical in the area propositions that follow (Props 35-48), where Euclid compares areas without using numerical measurements — a remarkable feat of geometric reasoning.

Discussion Questions

The proof uses the diagonal to split the parallelogram into two congruent triangles. A parallelogram has two diagonals. Does the proof work the same way with the other diagonal?
Euclid proves three properties in one proposition. Which of the three (equal opposite sides, equal opposite angles, diagonal bisects area) is most fundamental? Could you derive the other two from just one?
The converse question: if a quadrilateral has equal opposite sides, is it necessarily a parallelogram? What about equal opposite angles? Which conditions are sufficient?

Euclid's Original Proof

In parallelogrammic areas the opposite sides and angles are equal to one another, and the diameter bisects the areas.

Let ACDB be a parallelogrammic area, and BC its diameter;

I say that the opposite sides and angles of the parallelogram ACDB are equal to one another, and the diameter BC bisects it.

For, since AB is parallel to CD, and the straight line BC has fallen upon them,

the alternate angles ABC, BCD are equal to one another. [I.29]

Again, since AC is parallel to BD, and BC has fallen upon them,

the alternate angles ACB, CBD are equal to one another. [I.29]

Therefore ABC, DCB are two triangles having the two angles ABC, BCA equal to the two angles DCB, CBD respectively, and one side equal to one side, namely that adjoining the equal angles which is common to both, namely BC;

therefore they will also have the remaining sides equal to the remaining sides respectively, and the remaining angle to the remaining angle; [I.26]

therefore the side AB is equal to CD, and AC to BD, and further the angle BAC is equal to the angle BDC. [I.26]

And, since the angle ABC is equal to the angle BCD,

and the angle CBD to the angle ACB,

the whole angle ABD is equal to the whole angle ACD. [C.N. 2]

And the angle BAC was also proved equal to the angle BDC.

Therefore in parallelogrammic areas the opposite sides and angles are equal to one another.

I say, next, that the diameter also bisects the areas.

For, since AB is equal to CD, and BC is common,

the two sides AB, BC are equal to the two sides DC, CB respectively;

and the angle ABC is equal to the angle BCD;

therefore the base AC is also equal to the base DB,

and the triangle ABC is equal to the triangle DCB. [I.4]

Therefore the diameter BC bisects the parallelogram ACDB.

Q.E.D.

What's Next

Now that we know a diagonal bisects a parallelogram into two equal triangles, Euclid is ready to compare areas of parallelograms that aren't congruent at all. Proposition 35 delivers the surprising result that any two parallelograms on the same base and between the same parallel lines have equal area, even if they look completely different.