The congruence theorems SSS and SSA

The congruence theorem side, side, side (SSS): if the three sides of one triangle are equal to the three sides of another triangle, the triangles are congruent.

• In the drawing: $\displaystyle {AB}={DE}$, $\displaystyle {BC}={EF}$ and $\displaystyle {CA}={FD}$, so $\displaystyle \triangle ABC\cong\triangle DEF$.
• Why this is enough: we draw $\displaystyle BC$. The vertex $\displaystyle A$ is at the distance $\displaystyle BA$ from $\displaystyle B$ – on a circle about $\displaystyle B$ – and at the distance $\displaystyle CA$ from $\displaystyle C$ – on a circle about $\displaystyle C$. The two circles meet at one point above $\displaystyle BC$ (and at one more, its reflection, below it).
• So the lengths of the sides fix the angles too: a triangle cannot change an angle without changing a side.

The congruence theorem side, side, angle (SSA): if two sides, and the angle opposite the longer of them, in one triangle are equal to two sides, and the angle opposite the longer of them, in another triangle, the triangles are congruent.

• In the drawing: $\displaystyle {AB}={DE}$ and $\displaystyle {AC}={DF}$, and the side $\displaystyle AC$ is longer than $\displaystyle AB$.
• The angle opposite $\displaystyle AC$ is $\displaystyle \angle B$, and $\displaystyle {\angle B}={\angle E}$. So $\displaystyle \triangle ABC\cong\triangle DEF$.
• An important case: in a right-angled triangle the hypotenuse is opposite the right angle, and it is the longest side. So two right-angled triangles with equal hypotenuses and one pair of equal legs are congruent.

When the angle is opposite the shorter of the two sides, three equal parts are not enough.
In the drawing, in triangles $\displaystyle ABC$ and $\displaystyle ABC'$:
$\displaystyle AB$ is a common side, $\displaystyle {BC}={BC'}$, and the angle $\displaystyle A$ is common – and it is opposite $\displaystyle BC$, the shorter side.

• The triangles are not congruent: $\displaystyle AC'$ is much longer than $\displaystyle AC$.
• Why: we draw $\displaystyle \angle A$, and $\displaystyle AB$ along one arm. The third vertex is on the other arm, at the distance $\displaystyle BC$ from $\displaystyle B$ – on the circle about $\displaystyle B$. The circle cuts the other arm twice: at $\displaystyle C$ and at $\displaystyle C'$.
• When the angle is opposite the longer side, the circle cuts the other arm only once, so there is only one triangle – this is the theorem SSA.

Two triangles are congruent when they have one of these four sets of equal parts:

• SAS – two sides and the angle between them.
• ASA – a side and the two angles beside it.
• SSS – three sides.
• SSA – two sides and the angle opposite the longer of them.

Three equal angles are not enough: an equilateral triangle with a side of 1 cm and one with a side of 5 cm have the same angles, $\displaystyle 60^{\circ}$ each.

Are triangles $\displaystyle ABC$ and $\displaystyle DEF$ always congruent?
If so, write the congruence theorem. If not, give a reason.

$\displaystyle {AB}={DE}={5}$, $\displaystyle {BC}={EF}={7}$, $\displaystyle {CA}={FD}={6}$.

$\displaystyle {AB}={DE}={6}$, $\displaystyle {AC}={DF}={8}$, $\displaystyle {\angle B}={\angle E}={70^{\circ}}$.

$\displaystyle {AB}={DE}={8}$, $\displaystyle {AC}={DF}={6}$, $\displaystyle {\angle B}={\angle E}={40^{\circ}}$.

$\displaystyle {\angle A}={\angle D}={50^{\circ}}$, $\displaystyle {\angle B}={\angle E}={60^{\circ}}$, $\displaystyle {\angle C}={\angle F}={70^{\circ}}$.

In the quadrilateral $\displaystyle ABCD$: $\displaystyle {AB}={AD}$, and $\displaystyle {CB}={CD}$.
$\displaystyle AC$ is a diagonal.
Prove: $\displaystyle \triangle ABC\cong\triangle ADC$.

In triangles $\displaystyle ABC$ and $\displaystyle DEF$: $\displaystyle {\angle C}={\angle F}={90^{\circ}}$.
$\displaystyle {AB}={DE}$, and $\displaystyle {AC}={DF}$.
Prove: $\displaystyle \triangle ABC\cong\triangle DEF$.