Congruent Triangles
Two triangles are congruent when they are exactly the same shape and size, written \(\triangle ABC\equiv\triangle DEF\) with the vertices in matching order. Four tests prove it: SSS, SAS, AAS and RHS. In NSW Year 10 Advanced maths you choose the right test, use congruence to find unknown sides and angles, and write formal proofs (NESA outcome MA5‑GEO‑P‑01).
You will name the congruence test for pairs of triangles, spot a wrong vertex order, and find which extra fact is not enough. You will use matching sides and angles to form and solve equations, including one that leads to a quadratic. Proofs show that an isosceles triangle has equal base angles, that a quadrilateral with equal opposite sides is a parallelogram, and that a point on an angle bisector is the same distance from both arms. It is part of Year 10 Advanced Geometric Figures.
Every question with a fully worked solution.
- Congruent Triangles - Video - Congruent Triangle Rules Watch
Theory
Two triangles are congruent when one fits exactly on the other. Four tests prove it: SSS, SAS, AAS and RHS. This Year 10 Advanced maths page covers congruence statements in matching order, finding unknown parts, spotting when there is no test, and writing formal proofs. It comes up if you are heading for Extension maths.
Two figures are congruent if a translation, reflection or rotation moves one exactly onto the other. All matching sides and all matching angles are equal. The symbol is \(\equiv\).
Four tests prove two triangles congruent. SSS: three pairs of equal sides. SAS: two pairs of equal sides and the included angle between them. AAS: two pairs of equal angles and one pair of matching sides. RHS: a right angle, equal hypotenuses and one other pair of equal sides.
In a congruence statement such as \(\triangle ABC \equiv \triangle PQR\), the vertices are written in matching order, so \(AB = PQ\), \(\angle B = \angle Q\), and so on.
Three equal angles (AAA) and two sides with an angle that is not between them (SSA) are not tests.
This topic comes up if you are heading for Extension maths in Years 11 and 12.
| Test | You need | Watch for |
|---|---|---|
| SSS | three pairs of equal sides | a common side counts once for each triangle |
| SAS | two sides and the angle between them | the angle must be included |
| AAS | two angles and a matching side | the sides must be opposite equal angles |
| RHS | right angle, hypotenuse, one side | the hypotenuse is opposite the right angle |
How to write a congruence proof
- Name the two triangles: "In \(\triangle ABD\) and \(\triangle CDB\)".
- List three facts, one per line, each with a reason (given, common side, vertically opposite angles, alternate angles with the parallel lines named, and so on).
- Conclude with the congruence statement in matching order and the test in brackets.
- Hence: use "matching sides (or angles) of congruent triangles" to prove the extra fact the question asks for.
Other question types
- Finding parts: pair up the vertices from the statement, then copy the matching side or angle. Use the angle sum if the matching angle is not given.
- Algebra: set matching parts equal. Two pairs give simultaneous equations; a quadratic may give a value to reject because another pair of sides would not match.
- Choosing the test: mark every equal part on the diagram first, including the hidden ones.
- Lengths after congruence: often Pythagoras finishes the job, for example a peg distance or a kite's sides.
- Properties of shapes: diagonals of a parallelogram bisect each other, a point on an angle bisector is the same distance from both arms, and a point on the perpendicular bisector of \(AB\) is the same distance from \(A\) and \(B\). Each is proved with congruent triangles.
(i) Find \(TS\) and \(\angle T\).
(ii) \(RT = (4p - 3)\) cm. Find \(p\).
Read the matching pairs from the order of the letters.
| \(K \leftrightarrow R,\ L \leftrightarrow T,\ M\) | \(\leftrightarrow\) | \(S\) |
(i) \(TS\) matches \(LM\), so \(TS = \mathbf{14}\) cm. \(\angle T\) matches \(\angle L\).
| \(\angle L\) | \(=\) | \(180^\circ - 48^\circ - 71^\circ = 61^\circ\) |
| \(\angle T\) | \(=\) | \(\mathbf{61^\circ}\) |
(ii) \(RT\) matches \(KL\).
| \(4p - 3\) | \(=\) | \(9\) |
| \(p\) | \(=\) | \(\mathbf{3}\) |
Matching sides join matching vertices: \(TS\) joins the 2nd and 3rd letters, so it matches \(LM\), the 2nd and 3rd letters of the other triangle.
(i) \(AB = PQ = 5\), \(BC = QR = 7\), \(\angle C = \angle R = 40^\circ\)
(ii) \(\angle A = \angle P = 90^\circ\), \(BC = QR = 13\), \(AB = PQ = 5\)
(iii) \(\angle A = \angle P = 50^\circ\), \(\angle B = \angle Q = 60^\circ\), \(BC = QR\)
(iv) \(\angle A = \angle P\), \(\angle B = \angle Q\), \(\angle C = \angle R\)
For each, list the three facts and check whether they fit SSS, SAS, AAS or RHS.
(i) Two sides and an angle, but \(\angle C\) is not between \(AB\) and \(BC\) (the included angle would be \(\angle B\)). This is side-side-angle, which is not a test: two different triangles can fit it.
(ii) A right angle, the hypotenuses \(BC\) and \(QR\) (opposite the right angles) and one other side: congruent (RHS).
(iii) Two angles and a pair of matching sides. \(BC\) is opposite \(\angle A\) and \(QR\) is opposite \(\angle P\), so they match: congruent (AAS).
(iv) Three angles only. The triangles have the same shape but could be different sizes: not a test (they are similar, not necessarily congruent).
(i) Prove that \(\triangle AOC \equiv \triangle BOD\).
(ii) Hence prove that \(AC \parallel DB\).
(i) Name the two triangles, then give three facts with reasons.
\(AO = BO\) (\(O\) is the midpoint of \(AB\))
\(CO = DO\) (\(O\) is the midpoint of \(CD\))
\(\angle AOC = \angle BOD\) (vertically opposite angles)
\(\therefore \triangle AOC \equiv \triangle BOD\) (SAS)
(ii) Use a pair of matching angles from the congruent triangles.
\(\angle OAC = \angle OBD\) (matching angles of congruent triangles)
\(\therefore AC \parallel DB\) (alternate angles on transversal \(AB\) are equal)
The word "hence" means use part (i). Keep the vertices in matching order: \(A \leftrightarrow B\), \(O \leftrightarrow O\), \(C \leftrightarrow D\).
(i) Prove that \(\triangle ABE \equiv \triangle BCF\).
(ii) Hence find \(BF\), correct to two decimal places.
(i) In \(\triangle ABE\) and \(\triangle BCF\):
\(AB = BC\) (sides of a square)
\(\angle ABE = \angle BCF = 90^\circ\) (angles of a square)
\(BE = CF = 3\) (given)
\(\therefore \triangle ABE \equiv \triangle BCF\) (SAS)
(ii) \(BF\) matches \(AE\) (matching sides of congruent triangles). Find \(AE\) with Pythagoras.
| \(AE^2\) | \(=\) | \(8^2 + 3^2 = 73\) |
| \(BF = AE\) | \(=\) | \(\sqrt{73} \approx \mathbf{8.54}\text{ cm}\) |
The matching angles give more: \(\angle CBF = \angle BAE\), so \(\angle CBF + \angle BEA = \angle BAE + \angle BEA = 90^\circ\) (angle sum of \(\triangle ABE\)). The third angle of the small triangle where \(AE\) meets \(BF\) is therefore \(90^\circ\): the segments are perpendicular.
Common pitfalls
Frequently asked questions
What are the four tests for congruent triangles?
SSS (three pairs of equal sides), SAS (two sides and the included angle), AAS (two angles and a matching side) and RHS (right angle, hypotenuse and one other side).
Why is AAA not a congruence test?
Triangles with the same three angles have the same shape but can be different sizes. A small and a large equilateral triangle both have three \(60^\circ\) angles but are not congruent.
Why does the order of letters matter in a congruence statement?
The order shows which vertices match. If \(\triangle ABC \equiv \triangle PQR\), then \(A\) matches \(P\), \(B\) matches \(Q\) and \(C\) matches \(R\), so \(AB = PQ\) and \(\angle B = \angle Q\).
What is the difference between SAS and SSA?
In SAS the equal angle lies between the two equal sides, and it proves congruence. In SSA the angle is not between them, and two different triangles can be drawn, so it is not a test.
How do you write a congruence proof?
Name the two triangles, list three pairs of equal parts with a reason for each, then write the congruence statement in matching order with the test. Use matching parts of congruent triangles for any 'hence' part.