Triangle Side-Side-Side Congruence

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Theorem

Let two triangles have all $3$ sides equal.

Then they also have all $3$ angles equal.


Thus two triangles whose sides are all equal are themselves congruent.


In the words of Euclid:

If two triangles have the two sides equal to two sides respectively, and have also the base equal to the base, they will also have the angles equal which are contained by the equal straight lines.

(The Elements: Book $\text{I}$: Proposition $8$)


Proof 1

Euclid-I-8.png

Let $\triangle ABC$ and $\triangle DEF$ be two triangles such that:

$AB = DE$
$AC = DF$
$BC = EF$

Suppose $\triangle ABC$ were superimposed over $\triangle DEF$ so that point $B$ is placed on point $E$ and the side $BC$ on $EF$.

Then $C$ will coincide with $F$, as $BC = EF$ and so $BC$ coincides with $EF$.


Aiming for a contradiction, suppose $BA$ does not coincide with $ED$ and $AC$ does not coincide with $DF$.

Then they will fall as, for example, $EG$ and $GF$.

Thus there will be two pairs of straight line segments constructed on the same line segment, on the same side as it, meeting at different points.

But from Proposition $7$: Two Lines Meet at Unique Point, these points are coincident.

It follows by Proof by Contradiction that:

$BA$ coincides with $ED$
$AC$ coincides with $DF$.

Therefore $\angle BAC$ coincides with $\angle EDF$ and therefore:

$\angle BAC = \angle EDF$


The same argument can be applied mutatis mutandis to the other two sides.

Thus we have shown that all corresponding angles are equal.

$\blacksquare$


Proof 2

Let $\triangle ABC$ and $\triangle DEF$ have all three pairs of corresponding sides equal.

Let $AC$ be the longest side of $\triangle ABC$.

If there is more than one longest side, choose an arbitrary one.

We have by hypothesis:

$AC = DF$


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There are two cases:

case 1

The two triangles can be arranged as shown. They are reflections.

Side-Side-Side1.png


case 2

Otherwise, they are superimposable by the method in Triangle Side-Angle-Side Congruence).

In the words of Euclid (Euclid:Proposition/I/4:

"If the triangle ABC is superposed on the triangle DEF, and if the point A is placed on the point D and the straight line AB on DE, then the point B also coincides with E, because AB equals DE..."

If so, then reflect $\triangle DEF$ in $DF$.

Now arrange $\triangle ABC$ and $\triangle DEF$ with $AC = DF$ shared, as shown in the figure.


Construct $BE$.

We have by hypothesis:

$AB = AE$
$BC = EC$

By the definition of isosceles triangles:

$\triangle ABE$ and $\triangle BEC$ are both isosceles.

By Isosceles Triangle has Two Equal Angles:

$\angle ABE = \angle AEB$
$\angle EBC = \angle BEC$

By addition:

$\angle ABC = \angle AEC$

By Triangle Side-Angle-Side Congruence:

$\triangle ABC \cong \triangle DEF$

$\blacksquare$


Also known as

Triangle Side-Side-Side Congruence is also known as SSS or the SSS Condition.


Sources

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