Triangle Side-Angle-Angle Equality

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Theorem

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 congruent.


In the words of Euclid:

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 subtending one of the equal angles, they will also have the remaining sides equal to the remaining sides and the remaining angle equal to the remaining angle.

(The Elements: Book $\text{I}$: Proposition $26$)


Proof

Triangle SAA Equality.png

Let:

$\angle ABC = \angle DEF$
$\angle BCA = \angle EFD$
$AB = DE$

Aiming for a contradiction, suppose that $BC \ne EF$.

If this is the case, one of the two must be greater.

Without loss of generality, 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:

$BH = EF$
$\angle ABH = \angle DEF$
$AB = DE$

from Triangle Side-Angle-Side Equality we have:

$\angle BHA = \angle EFD$

But from External Angle of Triangle Greater than Internal Opposite, we have:

$\angle BHA > \angle HCA = \angle EFD$

which is a contradiction.

Therefore $BC = EF$.

So from Triangle Side-Angle-Side Equality:

$\triangle ABC = \triangle DEF$

$\blacksquare$


Historical Note

This theorem is the second part of Proposition $26$ of Book $\text{I}$ of Euclid's The Elements.


Sources