Congruence Criteria for Triangles for the ESAT

What defines congruent shapes?

In geometry, two shapes are described as congruent if they are identical in both shape and size. When two triangles are congruent, every side length and every internal angle in one triangle matches exactly with a corresponding side length and internal angle in the second triangle. However, you do not need to measure all six components to prove congruence: specific criteria allow you to prove it with just three.

The SSS Criterion (Side, Side, Side)

The SSS criterion states that two triangles, $A$ and $B$, are congruent if the three sides of triangle $A$ are equal in length to the three sides of triangle $B$.

The SAS Criterion (Two sides and the included angle)

The SAS criterion states that two triangles, $A$ and $B$, are congruent if two sides and the angle between those two sides (the included angle) in shape $A$ are the same as the two corresponding sides and the included angle in shape $B$.

The ASA Criterion (Two angles and a corresponding side)

The ASA criterion states that two triangles, $X$ and $Y$, are congruent if two angles of $X$ are the same size as two angles of $Y$, and a corresponding side of each triangle is equal.

In this context, the term 'corresponding' means that the sides are located in the same position relative to the angles. For example, if the side is opposite a specific angle in the first triangle, it must be opposite the same angle in the second triangle to count as corresponding.

In the diagrams below for triangles $X$ and $Y$, the equal sides correspond because they are both situated opposite the angle marked with a single arc.

The RHS Criterion (Right angle, hypotenuse, and side)

The RHS criterion is a special case for right-angled triangles. Two triangles, $A$ and $B$, are congruent if:

1. Both triangles contain a right angle ($90^\circ$).
2. They have the same length hypotenuse (the longest side, located opposite the right angle).
3. They have one other side of the same length.

Worked Example: Proving congruence with SSS

Consider triangle $PQR$, which is an isosceles triangle where $PQ = PR$. Let $S$ be the midpoint of the base $QR$. Show that triangle $PQS$ is congruent to triangle $PRS$.

First, always draw a diagram to visualise the problem.

We can establish congruence by looking at the three sides of triangles $PQS$ and $PRS$:

1. $PQ = PR$: This is given because the triangle is isosceles.
2. $PS$ is common: Both triangles share this side.
3. $QS = SR$: This is given by the fact that $S$ is the midpoint of $QR$.

Since all three pairs of corresponding sides are equal, triangle $PQS$ is congruent to triangle $PRS$ by the SSS criterion.

Worked Example: Identifying SAS congruence

Look at the three triangles below. Which two must be congruent? Note that these diagrams are not drawn to scale, so you must rely on the given notation rather than visual estimation.

All three triangles possess two sides of the same length. However, for the SAS criterion to apply, the angle must be 'included' (located between the two known sides).

In triangles $X$ and $Z$, the equal angles are included between the marked sides. In triangle $Y$, the marked angle is not between the two sides. Therefore, triangles $X$ and $Z$ are congruent.

Worked Example: Identifying ASA congruence

Examine the following three triangles to determine which two are congruent.

All three triangles have two equal angles. To prove congruence via ASA, we must ensure the equal side is in a corresponding position.

In triangle $X$, the side with the single bar is opposite the angle with one arc. In triangles $Y$ and $Z$, the side with the single bar is opposite the angle with two arcs. Because the sides in $Y$ and $Z$ are in the same position relative to the angles, triangles $Y$ and $Z$ are congruent.

Worked Example: Proving RHS congruence in a circle

$AB$ is the diameter of a circle $S$. Points $C$ and $D$ lie on the circumference on opposite sides of the diameter. If $AC = BD$, show that triangle $ACB$ is congruent to triangle $ADB$.

To prove this, we can use the RHS criterion:

1. Angle $ACB = 90^\circ$ and angle $ADB = 90^\circ$: This is a known geometric property where the angle subtended by a diameter at the circumference is always a right angle.
2. The hypotenuse of triangle $ACB$ is $AB$ and the hypotenuse of triangle $ADB$ is $AB$: They share the same hypotenuse, which is the diameter.
3. $AC = BD$: This is given in the problem description.

Since they have a right angle, an identical hypotenuse, and one other equal side, triangle $ACB$ is congruent to triangle $ADB$ by the RHS criterion.