Proposition 26 of 48 • Theorem

If two triangles have two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that opposite one of the equal angles, then the remaining sides equal the remaining sides and the remaining angle equals the remaining angle.

If two triangles share two equal angles and one equal side — whether that side lies between the equal angles (ASA) or opposite one of them (AAS) — then the triangles are completely congruent: all remaining sides and angles match.

Before You Read

You already know that two equal sides and the included angle (SAS) fully determine a triangle, and so do three equal sides (SSS). But what about angles? If two triangles share two equal angles and one equal side, must they be identical? Proposition 26 says yes—covering both the case where the equal side is between the equal angles (ASA) and the case where it is opposite one of them (AAS).

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What Euclid Is Doing

Setup: We have two triangles ABC and DEF with ∠ABC = ∠DEF and ∠BCA = ∠EFD. We consider two cases: (1) ASA — the included side BC = EF, or (2) AAS — a non-included side, say AB = DE (opposite one of the equal angles).

Approach: Both cases are proved by contradiction using the 'cut and match' technique. Assume a pair of supposedly equal sides are actually unequal. Cut the longer one down to match the shorter (Proposition 3), forming a new triangle. Then apply SAS (Proposition 4) to show this new triangle is congruent to the other, which forces an angle equality that contradicts the exterior angle theorem (Proposition 16).

Conclusion: Case 1 (ASA): Given ∠ABC = ∠DEF, BC = EF, ∠BCA = ∠EFD. Suppose AB is not equal to DE; say AB is greater. Cut off BG = DE on BA (Prop 3). Then in triangles GBC and DEF: BG = DE, BC = EF, ∠GBC = ∠DEF. By SAS (Prop 4), ∠GCB = ∠DFE = ∠BCA (given). But ∠GCB is only part of ∠BCA (since G is between A and B), and a part cannot equal the whole — contradiction (Common Notion 5). So AB = DE. Then by SAS (Prop 4), the triangles are fully congruent. Case 2 (AAS): Given ∠ABC = ∠DEF, ∠BCA = ∠EFD, BC = EF. Suppose AB is not equal to DE; say AB is greater. Cut off BG = DE on BA (Prop 3). In triangles GBC and DEF: BG = DE, BC = EF, ∠GBC = ∠DEF. By SAS (Prop 4), ∠BCG = ∠EFD. But ∠BCA = ∠EFD (given), so ∠BCG = ∠BCA. But ∠BCG is only part of ∠BCA (G lies between A and B on BA), so part = whole — contradiction (Common Notion 5). So AB = DE, and by SAS the triangles are congruent. ✓

Key Moves

Case 1 (ASA): Given ∠ABC = ∠DEF, BC = EF, ∠BCA = ∠EFD. Assume AB > DE.
Cut BG = DE on BA (Proposition 3). Apply SAS (Prop 4) to triangles GBC and DEF.
This gives ∠GCB = ∠DFE = ∠BCA, making an exterior angle equal to an interior opposite angle in triangle AGC — contradiction with Proposition 16.
So AB = DE. By SAS (Prop 4), the triangles are fully congruent ✓
Case 2 (AAS): Given ∠ABC = ∠DEF, ∠BCA = ∠EFD, BC = EF. Assume AB > DE.
Cut BG = DE on BA (Proposition 3). Apply SAS (Prop 4) to triangles GBC and DEF.
This gives ∠BCG = ∠EFD = ∠BCA — but ∠BCG is only part of ∠BCA (G between A and B), so part = whole. Contradiction with Common Notion 5.
So AB = DE, and by SAS the triangles are congruent ✓

Try It Yourself

On paper, draw a triangle and mark two of its angles. Now try to draw a second triangle with the same two angle measures and the same length of one specific side—but a different shape. You will find you cannot: the triangle is completely locked in. Try the ASA case (equal side between the equal angles) and the AAS case (equal side opposite one angle) separately.

Proof Challenge

Available Justifications

Case 1 (ASA): Assume AB ≠ DE; suppose AB > DE. Cut off BG = DE on AB

Drag justification

In triangles GBC and DEF: GB = DE, BC = EF, ∠GBC = ∠DEF, so △GBC ≅ △DEF by SAS

Drag justification

Then ∠GCB = ∠DFE = ∠ACB — but ∠GCB < ∠ACB (part < whole), contradiction

Drag justification

So AB = DE; by SAS (Prop 4), the triangles are fully congruent

Drag justification

Case 2 (AAS): Given ∠ABC = ∠DEF, ∠BCA = ∠EFD, BC = EF. Assume AB ≠ DE; suppose AB > DE. Cut off BG = DE on BA.

Drag justification

In triangles GBC and DEF: BG = DE, BC = EF, ∠GBC = ∠DEF. So △GBC ≅ △DEF by SAS.

Drag justification

Then ∠BCG = ∠EFD. But ∠BCA = ∠EFD (given). So ∠BCG = ∠BCA.

Drag justification

But ∠BCG is only part of ∠BCA (G lies between A and B on BA), so part = whole — contradiction. Therefore AB = DE, and by SAS the triangles are congruent.

Drag justification

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Curriculum Materials

Get the Teaching Materials

Structured lesson plan, student worksheet, and answer key for Proposition 26.

✓Lesson Plan (prop-26-lesson-plan.pdf)
✓Student Worksheet (prop-26-worksheet.pdf)
✓Answer Key (prop-26-answer-key.pdf)

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Why It Matters

This is the third and final triangle congruence theorem in Book I, covering both ASA and AAS. Together with SAS (Prop 4) and SSS (Prop 8), it completes the basic toolkit for proving triangles congruent. Notably, Euclid proves two theorems in one proposition — an unusual and efficient move.

Modern connection: ASA and AAS congruence are workhorses of surveying and navigation. If you can measure one side and two angles of a triangle (much easier than measuring all three sides across rough terrain), you can determine the entire triangle. This is the principle behind triangulation networks that mapped the globe.

Historical note: Proposition 26 is the last of Euclid's congruence theorems that does not depend on the parallel postulate. Everything from Prop 1 through Prop 26 belongs to what is called 'absolute geometry' — results that hold in both Euclidean and non-Euclidean geometries. The parallel postulate enters for the first time in Proposition 29.

Discussion Questions

Why does Euclid need proof by contradiction here? With SAS (Prop 4), SSS (Prop 8) could be proved directly. What makes the angle-based congruences harder?
There is one combination Euclid does not prove: SSA (two sides and an angle opposite one of them). Why not? Can you construct a counterexample showing SSA does not guarantee congruence?
Euclid proves both ASA and AAS in a single proposition. A modern textbook might prove AAS as a corollary of ASA plus the angle sum theorem (Prop 32). Why can Euclid not do this — what has he not yet proved at this point?

Euclid's Original Proof

If two triangles have the two angles equal to two angles respectively, and one side equal to one side, namely, either the side adjoining the equal angles, or that subtending one of the equal angles, they will also have the remaining sides equal to the remaining sides and the remaining angle to the remaining angle.

Let ABC, DEF be two triangles having the two angles ABC, BCA equal to the two angles DEF, EFD respectively, namely the angle ABC to the angle DEF, and the angle BCA to the angle EFD; and let one side also be equal to one side, first that adjoining the equal angles, namely BC to EF;

I say that the remaining sides will also be equal to the remaining sides respectively, namely AB to DE and AC to DF, and the remaining angle BAC is equal to the remaining angle EDF.

For, if AB is unequal to DE, one of them is greater.

Let AB be greater, and let BG be made equal to DE; [I.3] and let GC be joined. [Post. 1]

Then, since BG is equal to DE, and BC to EF, the two sides GB, BC are equal to the two sides DE, EF respectively; and the angle GBC is equal to the angle DEF;

therefore the base GC is equal to the base DF, and the triangle GBC is equal to the triangle DEF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I.4]

therefore the angle GCB is equal to the angle DFE.

But the angle DFE is by hypothesis equal to the angle BCA;

therefore the angle GCB is also equal to the angle BCA, [C.N. 1]

the less to the greater: which is impossible.

Therefore AB is not unequal to DE, and is therefore equal to it.

But BC is also equal to EF;

therefore the two sides AB, BC are equal to the two sides DE, EF respectively, and the angle ABC is equal to the angle DEF;

therefore the base AC is equal to the base DF, and the remaining angle BAC is equal to the remaining angle EDF. [I.4]

Again, let sides which subtend the equal angles be equal, as AB to DE;

I say again that the remaining sides will be equal to the remaining sides, namely BC to EF and AC to DF, and further the remaining angle BAC is equal to the remaining angle EDF.

For, if BC is unequal to EF, one of them is greater.

Let BC be greater, if possible, and let BH be made equal to EF; [I.3] and let AH be joined. [Post. 1]

Then, since BH is equal to EF, and AB to DE, the two sides AB, BH are equal to the two sides DE, EF respectively, and they contain equal angles;

therefore the base AH is equal to the base DF, and the triangle ABH is equal to the triangle DEF, and the remaining angles will be equal to the remaining angles, namely those which the equal sides subtend; [I.4]

therefore the angle BHA is equal to the angle EFD.

But the angle EFD is equal to the angle BCA;

therefore, in the triangle AHC, the exterior angle BHA is equal to the interior and opposite angle BCA:

which is impossible. [I.16]

Therefore BC is not unequal to EF, and is therefore equal to it.

But AB is also equal to DE;

therefore the two sides AB, BC are equal to the two sides DE, EF respectively, and they contain equal angles;

therefore the base AC is equal to the base DF, the triangle ABC is equal to the triangle DEF, and the remaining angle BAC is equal to the remaining angle EDF. [I.4]

Therefore etc.

Q.E.D.

What's Next

Proposition 26 completes the set of triangle congruence theorems—SAS, SSS, and now ASA/AAS—and closes what is called absolute geometry, results provable without the parallel postulate. Starting from Proposition 27, Euclid launches into the theory of parallel lines, where a transversal crossing two lines creates angle relationships that determine whether those lines are parallel.