SSS and RHS Congruence Criteria

If you know all three sides of a triangle, you know the triangle , no choice of shape remains. That is SSS. For right triangles, even less data is needed: the hypotenuse and one leg are enough. That is RHS. Both are essential tools for proofs involving triangles with no convenient angle information.

Definitions

SSS Criterion (Side–Side–Side). If the three sides of one triangle are respectively equal to the three sides of another, then the triangles are congruent.

RHS Criterion (Right angle–Hypotenuse–Side). If in two right triangles, the hypotenuse and one leg of one equal the hypotenuse and the corresponding leg of the other, then the triangles are congruent.

(RHS is sometimes called HL for "hypotenuse-leg".)

Why SSS works

Three side lengths uniquely determine the shape of a triangle (provided the triangle inequality is satisfied , see lesson 5). The intuition: fix the longest side as a base; the two shorter sides must reach a common third vertex. By the triangle inequality, exactly one such vertex exists above the base (and a mirror image below, which is the same triangle reflected). So three sides force the triangle up to a reflection , which is still considered congruent.

Why RHS works

Take two right triangles with equal hypotenuses and one pair of equal legs. By the Pythagoras theorem, the third side is forced: $\sqrt{H^2 - L^2}$. So all three sides match, and we have SSS.

In effect, RHS is SSS for right triangles, made simpler because one side is determined by the other two via Pythagoras. RHS is a separate named criterion just for convenience.

Writing an SSS proof

Example. In quadrilateral $ABCD$, $AB = CD$ and $AD = BC$. Prove $\triangle ABD \cong \triangle CDB$.

Proof.

By CPCT, $\angle ABD = \angle CDB$ and $\angle ADB = \angle CBD$. Q.E.D.

Writing an RHS proof

Example. $\triangle ABC$ and $\triangle DEF$ are both right-angled at $B$ and $E$ respectively, with $AC = DF$ (hypotenuses) and $AB = DE$ (one leg). Prove $\triangle ABC \cong \triangle DEF$.

Proof.

A classical application: perpendicular from the centre of a circle

Claim. A perpendicular dropped from the centre of a circle to a chord bisects the chord.

Proof sketch. Let $O$ be the centre and $AB$ a chord, with $M$ the foot of the perpendicular from $O$ to $AB$. Consider $\triangle OMA$ and $\triangle OMB$. Both are right-angled at $M$ (the perpendicular). $OA = OB$ (both radii). $OM = OM$ (common). By RHS, $\triangle OMA \cong \triangle OMB$. So $AM = MB$ by CPCT. The chord is bisected.

Worked examples

Example 1. $\triangle ABC$ has sides $5, 7, 9$. $\triangle DEF$ has sides $5, 7, 9$. Are they congruent?

Yes. By SSS, all three sides match.

Example 2. In a right triangle $\triangle ABC$ with right angle at $C$, $AB = 13$, $BC = 5$. In $\triangle DEF$ with right angle at $F$, $DE = 13$, $EF = 5$. Are they congruent?

Yes. Hypotenuse $AB = DE$, leg $BC = EF$, both right-angled. By RHS, congruent.

Example 3. $\triangle ABC$: sides $3, 4, 5$. $\triangle DEF$: sides $3, 4, 5$. Compute the angles using Pythagoras (you'll see $\angle C = 90^\circ$ in both, since $3^2 + 4^2 = 5^2$). By SSS the triangles are congruent.

Example 4. In a square $ABCD$, prove that the diagonals $AC$ and $BD$ bisect each other at right angles.

By symmetry of the square, $\triangle AOB \cong \triangle COD$ (with $O$ the intersection) and $\triangle BOC \cong \triangle AOD$ , by SSS (all sides equal). So $AO = OC$, $BO = OD$, and the angles at $O$ are equal, making them right angles by symmetry.

Example 5. Two right triangles $\triangle ABC$ (right-angled at $C$) and $\triangle DEF$ (right-angled at $F$) have $AB = DE$ and $AC = DF$. Are they congruent?

Hypotenuses $AB$ and $DE$ match; leg $AC$ and $DF$ match; right angles match. By RHS, congruent.

Try it yourself

1. State the SSS criterion.
2. State the RHS criterion.
3. Why is RHS only for right triangles?
4. In a kite (two pairs of equal adjacent sides), prove that one of the diagonals divides it into two congruent triangles using SSS.
5. In a rhombus, prove that the diagonals bisect each other using SSS.
6. A perpendicular is dropped from the centre of a circle to a chord. Prove the chord is bisected.
7. Two right triangles have hypotenuse $10$ and one leg $6$. Are they necessarily congruent?
8. Why does RHS reduce to SSS via Pythagoras?
9. If two triangles have all three sides equal, can the angles still differ? Justify.
10. In $\triangle ABC$ and $\triangle DEF$, $AB = DE$, $BC = EF$, $CA = FD$. State the congruence and name the criterion.

Pitfalls / Insight

• RHS needs both triangles to be right-angled. Don't apply it where there is no right angle.
• For RHS, the matching leg must correspond. Don't mix up which leg of one triangle matches which of the other.
• SSS does not need angle information. That is its strength , sometimes no angles are given.

Insight. SSS, SAS, ASA/AAS, and RHS cover almost every congruence question you will meet. Identifying which criterion the figure offers is the entire game; once chosen, the three-line proof is mechanical and CPCT delivers the rest.