ASA, SSS and RHS

Two angles and the side between them: ASA

Draw a side $AB$ of length $7\ \text{cm}$. At $A$ draw a ray at $40^\circ$ to the side, and at $B$ draw a ray at $65^\circ$ on the same side. The two rays cross at one point, which we call $C$. You get exactly one triangle, whoever draws it. That is the ASA rule.

ASA rule. If two angles and the side between them in one triangle are equal to two angles and the side between them in another triangle, the triangles are congruent.

The third angle gives AAS

The angles of every triangle add up to $180^\circ$. So if two angles of one triangle equal two angles of another, the third angles are equal too.

This turns a side that is not between the two angles into one that is. Suppose triangles $ABC$ and $PQR$ have $\angle B = \angle Q = 52^\circ$, $\angle C = \angle R = 71^\circ$ and $AB = PQ$. Then $\angle A = 57^\circ$ and $\angle P = 57^\circ$. The side $AB$ lies between the angles $A$ and $B$, and $PQ$ lies between $P$ and $Q$. That is ASA.

AAS rule. If two angles and a side that is not between them in one triangle are equal to the matching parts of another triangle, the triangles are congruent.

Example 1. The segments $AB$ and $CD$ are parallel, and $O$ is the middle point of $AD$. Show that $\triangle AOB \cong \triangle DOC$, and that $O$ is also the middle point of $BC$.

1. $\angle OAB = \angle ODC$. They are alternate angles, since $AB$ is parallel to $CD$.
2. $\angle AOB = \angle DOC$. They are vertically opposite angles.
3. $AO = DO$. The point $O$ is the middle of $AD$.
4. So $\triangle AOB \cong \triangle DOC$, by AAS.
5. Therefore $OB = OC$, by CPCT. So $O$ is the middle point of $BC$.

All three sides: SSS

Make a triangle from three sticks of different lengths, pinned at the corners. It cannot change shape. Push as you like, you can only move the whole frame. A four-sided frame is different: it leans over into a slanted shape. This is why a gate, a roof truss and a bridge have diagonal bars. A diagonal splits a four-sided frame into two triangles, and triangles do not lean.

SSS rule. If the three sides of one triangle are equal to the three sides of another triangle, the triangles are congruent.

Example 2. In a kite-shaped field $ABCD$, $AB = AD$ and $CB = CD$. Show that $\angle B = \angle D$.

1. $AB = AD$ and $CB = CD$. These are given.
2. $AC = AC$. This is the common side.
3. So $\triangle ABC \cong \triangle ADC$, by SSS.
4. Therefore $\angle B = \angle D$, by CPCT.

A right angle, the hypotenuse and a side: RHS

In a right triangle the longest side, opposite the right angle, is the hypotenuse. R, H and S stand for right angle, hypotenuse, side.

RHS rule. If two right triangles have equal hypotenuses and one other side equal, the triangles are congruent.

Example 3. The lines $BE$ and $CF$ are the altitudes (perpendiculars) from $B$ and $C$ to the opposite sides of triangle $ABC$, and $BE = CF$. Show that $\triangle BEC \cong \triangle CFB$.

1. $\angle BEC = \angle CFB = 90^\circ$. The altitudes meet the sides at right angles.
2. $BC = CB$. This is the common hypotenuse.
3. $BE = CF$. This is given.
4. So $\triangle BEC \cong \triangle CFB$, by RHS.

Three facts that are not enough

AAA does not work. Take a triangle with angles $40^\circ$, $60^\circ$ and $80^\circ$, and make a bigger copy on a photocopier. Every angle is the same, but the sizes differ. Three equal angles fix the shape, not the size.

SSA does not work. Take two sides and an angle that is not between them. The next picture shows that you can get two different triangles. Here $AB = 7\ \text{cm}$, $\angle A = 35^\circ$ and the second side is $6\ \text{cm}$ long. A circle of that radius around $B$ cuts the ray from $A$ in two places.

When the side from $B$ is at least as long as $AB$, only one point is left on the ray, and SSA works. A right angle is the most important case: the hypotenuse is always the longest side. That is why RHS is a rule.

Choosing the rule

Look at what you are given, and ask where the equal parts sit.

• Two sides and an equal angle between them: SAS.
• Two equal angles and the side between them: ASA. If the side is elsewhere, use the angle sum first, then ASA (AAS).
• Three equal sides: SSS.
• Right angles, equal hypotenuses and one more equal side: RHS.
• Only angles, or two sides and an angle elsewhere: not enough.

Practise with new numbers

Try it in Python

This program tries several lengths for the second side in the SSA situation, and counts the triangles that fit.

Common mistakes

• Using AAA. Three equal angles give triangles of the same shape, not of the same size.
• Using SSA. Two sides and an angle that is not between them can give two triangles.
• Using RHS without a right angle. The right angle must be in both triangles.
• Matching the wrong corners. Write the congruence in the order of the corners that match, and check the order again before you use CPCT.

Remember these

• ASA: two angles and the side between them. AAS: two angles and a side elsewhere.
• SSS: three sides. A triangle is rigid, so three sides fix it.
• RHS: a right angle, the hypotenuse and one side.
• AAA and SSA are not rules.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Triangles $ABC$ and $DEF$ have $\angle A = \angle D$, $\angle B = \angle E$ and $BC = EF$. Which rule can you use?

2. Why is a four-sided frame of sticks not rigid, while a triangle is?

3. Two triangles have angles $30^\circ$, $60^\circ$ and $90^\circ$. Must they be congruent?

4. In two right triangles, the hypotenuses are both $13\ \text{cm}$ and one side is $5\ \text{cm}$ in each. Are they congruent?

5. Sana draws a triangle with sides $6\ \text{cm}$ and $4\ \text{cm}$ and an angle of $30^\circ$ at the end of the $4\ \text{cm}$ side, away from the $6\ \text{cm}$ side. Is the triangle fixed?