# Geometry for Elementary School/The Side-Side-Side congruence theorem

The first congruence theorem we will discuss is the Side-Side-Side theorem.

## The Side-Side-Side congruence theorem

Given two triangles $\triangle ABC$ and $\triangle DEF$ such that their sides are equal, hence:
* The side ${\overline {AB}}$ equals ${\overline {DE}}$.
* The side ${\overline {BC}}$ equals ${\overline {EF}}$.
* The side ${\overline {AC}}$ equals ${\overline {DF}}$.
Then the triangles are congruent and their angles are equal too.

## Method of Proof

In order to prove the theorem we need a new postulate. The postulate is that one can move or flip any shape in the plane without changing it. In particular, one can move a triangle without changing its sides or angles. Note that this postulate is true in plane geometry but not in general. If one considers geometry over a ball, the postulate is no longer true.
Given the postulate, we will show how can we move one triangle to the other triangle location and show that they coincide. Due to that, the triangles are equal.

### The construction

* Copy The line Segment side ${\overline {AB}}$ to the point D.
* Draw the circle $\circ D,{\overline {AB}}$.
* The circle $\circ D,{\overline {AB}}$ and the segment ${\overline {DE}}$ intersect at the point E hence we have a copy of ${\overline {AB}}$ such that it coincides with ${\overline {DE}}$.
* Construct a triangle with ${\overline {DE}}$ as its base, ${\overline {BC}}$, ${\overline {AC}}$ as the sides and the vertex at the side of the vertex F. Call this triangle triangles $\triangle DEG$

### The claim

The triangles $\triangle DEF$ and $\triangle ABC$ congruent.

### The proof

* The points A and D coincide.
* The points B and E coincide.
* The vertex F is an intersection point of $\circ D,{\overline {DF}}$ and $\circ E,{\overline {EF}}$.
* The vertex G is an intersection point of $\circ D,{\overline {AC}}$ and $\circ E,{\overline {BC}}$.
* It is given that ${\overline {DF}}$ equals ${\overline {AC}}$.
* It is given that ${\overline {EF}}$ equals ${\overline {BC}}$.
* Therefore, $\circ D,{\overline {DF}}$ equals $\circ D,{\overline {AC}}$and $\circ E,{\overline {EF}}$ equals $\circ E,{\overline {BC}}$.
* However, circles of different centers have at most one intersection point in one side of the segment that joins their centers.
* Hence, the points G and F coincide.
* There is only a single straight line between two points, therefore ${\overline {EG}}$ coincides with ${\overline {EF}}$ and ${\overline {GD}}$ coincides with ${\overline {DF}}$.
* Therefore, the $\triangle DEG$ coincides with $\triangle DEF$ and the two are congruent.
* Due to the postulate $\triangle DEG$ and $\triangle ABC$ are equal and therefore congruent.
* Hence, $\triangle DEF$ and $\triangle ABC$ are congruent.
* Hence, $\angle ABC$ equals $\angle DEF$, $\angle BCA$ equals $\angle EFD$ and $\angle CAB$ equals $\angle FDE$.

## Note

The Side-Side-Side congruence theorem appears as Book I, prop 8 at the Elements. The proof here is in the spirit of the original proof. In the original proof Euclid claims that the vertices F and G must coincide but doesn’t show why. We used the assumption that “circles of different centers have at most one intersection point in one side of a segment that joins their centers”. This assumption is true in plane geometry but doesn’t follows from Euclid’s original postulates. Since Euclid himself had to use such an assumption, we preferred to give a more detailed proof, though the extra assumption.