The congruence theorems SAS and ASA

To know that two triangles are congruent, you need not check all six pairs of corresponding parts: some three parts fix the whole triangle. Every triangle built from three such parts comes out the same triangle – so two triangles that share those three parts are congruent.

• A congruence theorem says which three parts are enough.
• Its name is made of S (side) and A (angle), in the order the parts sit in the triangle.

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

• In the drawing: $\displaystyle {AB}={DE}$, $\displaystyle {\angle B}={\angle E}$ and $\displaystyle {BC}={EF}$, so $\displaystyle \triangle ABC\cong\triangle DEF$.
• The angle must be between the two sides: $\displaystyle \angle B$ is between $\displaystyle AB$ and $\displaystyle BC$.
• Why this is enough: we build the triangle. We draw $\displaystyle \angle B$, and measure $\displaystyle BA$ along one arm and $\displaystyle BC$ along the other. That fixes the vertices $\displaystyle A$ and $\displaystyle C$, and only one segment joins them – no choice is left.

The congruence theorem angle, side, angle (ASA): if a side and the two angles beside it in one triangle are equal to a side and the two angles beside it in another triangle, the triangles are congruent.

• In the drawing: $\displaystyle {\angle B}={\angle E}$, $\displaystyle {BC}={EF}$ and $\displaystyle {\angle C}={\angle F}$, so $\displaystyle \triangle ABC\cong\triangle DEF$.
• Why this is enough: we draw $\displaystyle BC$, and the two angles at its ends. The arm from $\displaystyle B$ and the arm from $\displaystyle C$ meet at one point only – the vertex $\displaystyle A$.

The segments $\displaystyle AB$ and $\displaystyle CD$ cross at the point $\displaystyle O$.
$\displaystyle O$ is the midpoint of $\displaystyle AB$, and $\displaystyle O$ is the midpoint of $\displaystyle CD$.
Prove: $\displaystyle \triangle AOC\cong\triangle BOD$.

The proof below finds two pairs of equal sides in triangles $\displaystyle AOC$ and $\displaystyle BOD$, and the angle between them – the angles at $\displaystyle O$ are vertical angles:

A proof uses equal parts that the data does not list; they are found in the drawing:

• A common side – a side of both triangles: it equals itself, $\displaystyle {AC}={AC}$.
• Vertical angles – they are equal.
• Alternate or corresponding angles between parallel lines – they are equal.

The triangles are congruent. Write the congruence in the right order, and the congruence theorem.

In triangles $\displaystyle ABC$ and $\displaystyle DEF$:
$\displaystyle {\angle A}={\angle D}={50^{\circ}}$, $\displaystyle {\angle B}={\angle E}={70^{\circ}}$.
$\displaystyle {BC}={EF}$.

$\displaystyle D$ is a point on the side $\displaystyle BC$ of triangle $\displaystyle ABC$.
$\displaystyle AD$ bisects the angle $\displaystyle A$: $\displaystyle {\angle BAD}={\angle CAD}$.
$\displaystyle AD\perp BC$.
Prove: $\displaystyle \triangle ABD\cong\triangle ACD$.

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