# Triangle Angle-Side-Angle and Side-Angle-Angle Equality

## Theorem

### Part 1

If two triangles have

• Two angles equal to two angles, respectively;
• The sides between the two angles equal

Then the remaining angles are equal, and the remaining sides equal the respective sides.

That is to say, if two pairs of angles and the included sides are equal, then the triangles are equal.

### Part 2

If two triangles have

• Two angles equal to two angles, respectively;
• The sides opposite one pair of equal angles equal

Then the remaining angles are equal, and the remaining sides equal the respective sides.

That is to say, if two pairs of angles and a pair of opposite sides are equal, then the triangles are equal.

## Proof

### Part 1

Let $\angle ABC = \angle DEF$, $\angle BCA = \angle EFD$, and $BC = EF$.

Assume $AB \neq DE$. If this is the case, one of the two must be greater. WLOG, we let $AB > DE$.

We construct a point $G$ on $AB$ such that $BG = ED$, and then we construct the segment $CG$.

Now, since we have $BG = ED$, $\angle GBC = \angle DEF$, and $BC = EF$, from Triangle Side-Angle-Side Equality we have $\angle GCB = \angle DFE$.

But from Euclid's fifth common notion $\angle DFE = \angle ACB > \angle GCB$, a contradiction.

Therefore, $AB = DE$, so from Triangle Side-Angle-Side Equality, we have $\triangle ABC = \triangle DEF$.

$\blacksquare$

### Part 2

Let $\angle ABC = \angle DEF$, $\angle BCA = \angle EFD$, and $AB = DE$.

Assume $BC \neq EF$. If this is the case, one of the two must be greater. WLOG, we let $BC > EF$.

We construct a point $H$ on $BC$ such that $BH = EF$, and then we construct the segment $AH$.

Now, since we have $BH = EF$, $\angle ABH = \angle DEF$, and $AB = DE$, from Triangle Side-Angle-Side Equality we have $\angle BHA = \angle EFD$.

But from External Angle of Triangle Greater than Internal Opposite, we have $\angle BHA > \angle HCA = \angle EFD$, a contradiction.

Therefore, $BC = EF$, so from Triangle Side-Angle-Side Equality, we have $\triangle ABC = \triangle DEF$.

$\blacksquare$

## Alternate Proof

Either part of this theorem follows trivially from the other part and the fact that the sum of the angles of a triangle equals two right angles. However, it is important to note that both of these are provable without the parallel postulate, which the proof of that theorem requires.

## Historical Note

This is Proposition 26 of Book I of Euclid's The Elements.