Two triangles are congruent when they are identical in every way. Master the four tests -- SSS, SAS, AAS and RHS -- and prove triangles match perfectly.
Today's hook: Two triangles both have sides 5 cm, 6 cm and 7 cm. Are they definitely identical? What if two triangles both have angles 40°, 60° and 80° -- are they definitely identical? Your intuition here is the key to everything.
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Two triangles both have sides 5 cm, 6 cm and 7 cm. Are they definitely identical? What if two triangles both have angles 40°, 60° and 80° -- are they definitely identical? Explain your reasoning.
Record your answer in your workbook.
1
The Big Idea
+5 XP to read
Congruence means identical in every way -- same shape, same size, same angles, same side lengths. If two triangles are congruent, every corresponding part matches. There are exactly four ways to prove this, and one famous way that does not work.
An SSS proof uses three sides. An SAS proof needs the angle between the two sides. AAS uses two angles and any corresponding side. RHS is the special right-triangle case that SSA is not.
$\triangle ABC \equiv \triangle DEF$ (test)
Four tests only
SSS, SAS, AAS, RHS. Memorise them all -- no substitutes.
SSA never works
Same two sides and a non-included angle can make two different triangles.
What congruence means for corresponding sides and angles
Why SSA is not a valid test
Understand
Why the included angle matters for SAS
How congruence proofs are structured step by step
Why matching vertices is essential in notation
Can Do
Identify which congruence test applies to any pair of triangles
Write formal congruence proofs with clear reasoning
Use congruence to deduce unknown sides and angles
3
Words You Need
vocabulary
CongruentIdentical in shape and size. Corresponding sides and angles are equal. Symbol: $\equiv$.
SSSSide-Side-Side. Three pairs of corresponding sides equal.
SASSide-Angle-Side. Two sides and the included angle equal.
AASAngle-Angle-Side. Two angles and a corresponding side equal.
RHSRight angle-Hypotenuse-Side. Right triangle with hypotenuse and one leg equal.
CorrespondingMatching parts in the same relative position in congruent figures.
4
Spot the Trap
heads-up
Wrong: Using SSA as a congruence test. Two different triangles can have the same two sides and a non-included angle.
Right: SAS requires the angle to be included between the two sides. SSA is never a valid test.
Wrong: Writing the congruence statement in the wrong order, so corresponding parts do not match.
Right: Always match vertices: if $\triangle ABC \equiv \triangle DEF$, then $A$ matches $D$, $B$ matches $E$, $C$ matches $F$.
5
The Four Tests
+5 XP
These are the only four ways to prove two triangles are identical. Memorise them all. Each test specifies exactly which matching parts force the triangles to be congruent.
SSS is the most straightforward -- match all three sides and the triangles must be identical. SAS is powerful but needs the angle between the sides. AAS works because two angles fix the shape, and one side fixes the size. RHS is the only case where SSA is valid -- because the right angle removes the ambiguity.
SSS | SAS | AAS | RHS
SSS = rigid triangle
A triangle frame with fixed side lengths cannot change shape.
SAS = hinge
Two sticks joined at a fixed angle make one rigid shape.
AAS = shape + size
Two angles fix the shape. One side fixes how large it is.
6
Why SSA Fails
+5 XP
Two triangles can have the same two sides and a non-included angle, but still be different shapes. This is called the ambiguous case. The only exception is RHS, where the right angle constraint removes the ambiguity.
Imagine a side of 5 cm, a side of 7 cm, and an angle of 30° that is not between them. You can draw two completely different triangles. The 30° angle could open to the left or the right, creating two possible positions for the third vertex.
SSA ≠ congruence
Angle must be included
For SAS, the angle sits between the two known sides.
RHS is the exception
The 90° angle and hypotenuse fix everything -- no ambiguity.
Test before you claim
Always verify the angle is included before calling it SAS.
7
Match the Vertices
+5 XP
The congruence statement $\triangle ABC \equiv \triangle DEF$ is not just shorthand -- it is a matching contract. $A$ corresponds to $D$, $B$ to $E$, $C$ to $F$. This means every corresponding side and angle must match in that exact order.
If $\triangle ABC \equiv \triangle DEF$ by SAS, then side $AB$ corresponds to side $DE$, side $BC$ to $EF$, and the included angle $\angle B$ corresponds to $\angle E$. Get the order wrong and your proof falls apart.
$AB = DE$, $BC = EF$, $AC = DF$
Name matching pairs
First vertex matches first, second matches second, third matches third.
Once you prove two triangles are congruent, you unlock every corresponding part. Unknown sides become known. Unknown angles become known. This is the real power of congruence -- it lets you transfer information from one triangle to another.
The standard pattern: prove congruence first using one of the four tests, then state that corresponding parts are equal (CPCTC -- Corresponding Parts of Congruent Triangles are Congruent). This second step often gives you the value you actually need.
CPCTC: congruent $\Rightarrow$ all parts match
Prove first, find second
Always establish congruence before claiming equal parts.
Label diagrams
Mark equal sides and angles as you spot them.
State the test
End every proof with the test name in brackets.
Watch Me Solve It · 3 examples
Watch Me Solve It · Identifying the Test
+15 XP per step
Q1
PROBLEM
In triangles $ABC$ and $DEF$, $AB = DE = 5$ cm, $BC = EF = 7$ cm, and $\angle B = \angle E = 50°$. Which congruence test applies?
1
List the given information
$AB = DE$ (one pair of sides), $BC = EF$ (second pair of sides), $\angle B = \angle E$ (one pair of angles)
Organise what you know before deciding on a test.
2
Check the position of the angle
The angle $\angle B$ is between sides $AB$ and $BC$. The angle $\angle E$ is between sides $DE$ and $EF$.
The angle is included between the two known sides -- this is the critical check.
3
Name the test
Two sides and the included angle are equal → SAS (Side-Angle-Side)
$\angle BAD = \angle CAD$ (corresponding angles of congruent triangles)
Once triangles are proven congruent, all corresponding parts are equal.
4
Substitute the known value
$\angle BAD = 35°$
Since $\angle CAD = 35°$ and the angles correspond, $\angle BAD = 35°$.
Nice work -- XP earned
Answer$\angle BAD = 35°$
9
Common Pitfalls
heads-up
Using SSA as a valid test
Writing "SSA" in a proof or selecting it in a multiple choice question. Two different triangles can share the same two sides and a non-included angle -- this is the ambiguous case.
Fix: remember the four valid tests -- SSS, SAS, AAS, RHS. If the angle is not included between the two sides, it is not SAS and it is not valid.
Writing vertices in the wrong order
Stating $\triangle ABC \equiv \triangle DEF$ when $A$ actually corresponds to $F$, not $D$. This makes every subsequent claim about corresponding parts incorrect.
Fix: trace the matching carefully. If $\angle A = \angle D$ and $\angle B = \angle E$, then $C$ must match $F$. Write the statement to reflect this.
Forgetting to state the test at the end
Writing "$\triangle ABC \equiv \triangle DEF$" without specifying which test justifies it. In formal proofs, the test name in brackets is required.
Fix: end every congruence proof with the test in brackets: (SSS), (SAS), (AAS), or (RHS).
Four quick problems -- identify which congruence test applies, or state "not possible." Work each one, then reveal the answer.
1 In triangles $PQR$ and $STU$, $\angle P = \angle S = 40°$, $\angle Q = \angle T = 70°$, and $QR = TU = 8$ cm. Which test?
AAS. Two angles and a corresponding side. Since two angles are equal, the third must also be equal, so we effectively know all angles and one corresponding side.$\triangle PQR \equiv \triangle STU$ (AAS)
2 Two right triangles have hypotenuse 13 cm and one leg 5 cm. Which test?
RHS. Right angle, hypotenuse and one side. This is the special case where SSA works because the right angle removes ambiguity.RHS
3 In triangles $XYZ$ and $LMN$, $XY = LM = 6$ cm, $YZ = MN = 9$ cm, and $\angle X = \angle L = 35°$. The angle is not included. Can we prove congruence?
No. This is SSA -- two sides and a non-included angle. Two different triangles could satisfy these conditions. Not a valid congruence test.Not possible
4 Triangle $ABC$ has sides $AB = 4$ cm, $BC = 5$ cm, $CA = 6$ cm. Triangle $DEF$ has $DE = 4$ cm, $EF = 5$ cm, $FD = 6$ cm. Which test?
SSS. All three pairs of corresponding sides are equal. The order given ($CA$ and $FD$) still matches because the sides correspond correctly.$\triangle ABC \equiv \triangle DEF$ (SSS)
Complete in your workbook.
Multiple Choice · 5 questions
MC1
All three sides
+10 XP
Which test proves two triangles congruent if all three sides of one equal the corresponding three sides of the other?
Correct -- SSS (Side-Side-Side) matches all three corresponding sides.
Not quite -- three sides means SSS. SAS needs an included angle, AAS needs angles, RHS needs a right triangle.
Explanation: SSS is the test where all three pairs of corresponding sides are equal. It is the most direct test and guarantees the triangles are identical.
MC2
Right triangle test
+10 XP
Two right-angled triangles have equal hypotenuses and one pair of equal legs. The correct test is:
Correct -- RHS (Right angle-Hypotenuse-Side) is the dedicated test for right triangles.
RHS is specifically for right triangles with matching hypotenuse and one leg.
Explanation: RHS requires a right angle, equal hypotenuses, and one pair of equal legs. While you could also use SSS (find the third side via Pythagoras), RHS is the direct test.
MC3
The ambiguous case
+10 XP
In triangles $ABC$ and $DEF$, $AB = DE$, $BC = EF$, and $\angle A = \angle D$. Can we prove congruence?
Correct -- this is SSA, not SAS, because the angle is not included between the two sides.
Check the position of the angle. SAS requires the angle to be between the two known sides.
Explanation: Here the known angle $\angle A$ is opposite side $BC$, not between $AB$ and $BC$. This is SSA (Side-Side-Angle), which is not a valid congruence test because two different triangles can satisfy these conditions.
MC4
Corresponding parts
+10 XP
If $\triangle ABC \equiv \triangle DEF$ by SAS, and $AB = 5$ cm, $BC = 7$ cm, then:
Correct -- $A \leftrightarrow D$, $B \leftrightarrow E$, $C \leftrightarrow F$. So $AB \leftrightarrow DE$ and $BC \leftrightarrow EF$.
Match the vertices carefully. $AB$ is the first two letters, so it corresponds to $DE$.
Explanation: In the statement $\triangle ABC \equiv \triangle DEF$, vertex $A$ corresponds to $D$, $B$ to $E$, and $C$ to $F$. Therefore side $AB$ (between $A$ and $B$) corresponds to side $DE$ (between $D$ and $E$), and side $BC$ corresponds to $EF$.
MC5
Angles and one side
+10 XP
Two triangles have angles $50°$, $60°$, $70°$ and side 8 cm opposite the $50°$ angle. They are:
Correct -- AAS needs the corresponding side to be in matching position. Same angles alone only guarantee similarity.
Two triangles with the same three angles are similar, not necessarily congruent. For congruence, a corresponding side must also match.
Explanation: Equal angles mean the triangles have the same shape (they are similar), but they could be different sizes. For AAS congruence, the given side must be a corresponding side -- opposite the same angle or between the same two angles -- in both triangles.
Short Answer · 3 questions
Q6
Isosceles triangle proof
+15 XP
Q6
SHORT ANSWER
In the diagram, $AB = AC$ and $BD = CD$. (a) Prove that $\triangle ABD \equiv \triangle ACD$. (b) Hence, prove that $\angle BAD = \angle CAD$. (c) What does this tell you about the line $AD$?
Write your working in your book.
(a) In triangles $ABD$ and $ACD$: $AB = AC$ (given), $BD = CD$ (given), $AD = AD$ (common). Therefore $\triangle ABD \equiv \triangle ACD$ (SSS).
(b) Since the triangles are congruent, corresponding angles are equal. Therefore $\angle BAD = \angle CAD$.
(c) $AD$ bisects $\angle BAC$. It is also the perpendicular bisector of $BC$ and the line of symmetry of the isosceles triangle $ABC$.
Marking guidance: 1 mark for (a) with reasoning, 1 mark for (b), 1 mark for (c).
Q7
Rectangle diagonals
+15 XP
Q7
SHORT ANSWER
$ABCD$ is a rectangle. The diagonals $AC$ and $BD$ intersect at point $E$. (a) Prove that $\triangle ABE \equiv \triangle CDE$. (b) Prove that the diagonals of a rectangle are equal in length. (c) Explain why this proof would not work for a general parallelogram.
(b) In triangles $ABC$ and $BAD$: $AB = BA$ (common), $BC = AD$ (opposite sides of rectangle), $\angle ABC = \angle BAD = 90°$. Therefore $\triangle ABC \equiv \triangle BAD$ (SAS), so $AC = BD$.
(c) In a general parallelogram, adjacent angles are not $90°$, so we cannot use SAS with a right angle. The alternate angles method still works but only proves the diagonals bisect each other, not that they are equal.
Marking guidance: 1 mark for (a), 2 marks for (b) with working, 1 mark for (c).
Q8
The 6-8-10 triangle
+15 XP
Q8
SHORT ANSWER
Two triangles have sides $a = 6$ cm, $b = 8$ cm, $c = 10$ cm and sides $d = 6$ cm, $e = 8$ cm with included angle $90°$. (a) Identify the congruence test that applies, if any. (b) Calculate the third side of the second triangle to verify. (c) A student claims SSA always works for right triangles. Is this true? Explain. (d) In what way does RHS differ from the general SSA case?
Write your working in your book.
(a) The first triangle is a 6-8-10 right triangle (since $6^2 + 8^2 = 10^2$). The second triangle has sides 6 and 8 with included angle $90°$. By Pythagoras, its third side is $\sqrt{6^2 + 8^2} = 10$ cm. Both triangles have sides 6, 8, 10 -- congruent by SSS (or by SAS).
(b) Third side $= \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$ cm.
(c)True for right triangles specifically. In a right triangle, if you know two sides, the third is fixed by Pythagoras, so SSA becomes equivalent to SSS. But this only works because the right angle constraint removes the ambiguity.
(d) In RHS, the known angle is always $90°$ and the known side is always opposite this angle (the hypotenuse). In general SSA, the angle is arbitrary and the side can be in any position, leading to two possible triangles. The right angle constraint means the third vertex must lie on a specific circle, and with the hypotenuse fixed, there is only one possible triangle.
Marking guidance: 1 mark for (a), 1 mark for (b), 1 mark for (c) with explanation, 2 marks for (d) with clear distinction.
S
Stretch Challenge · Overlapping triangles
+20 XP
S
STRETCH
In the figure, $ABCD$ is a square. $P$ is a point on $AB$ and $Q$ is a point on $AD$ such that $BP = DQ$. The lines $CP$ and $CQ$ are drawn. (a) Prove that $\triangle CBP \equiv \triangle CDQ$. (b) Hence prove that $\triangle CPQ$ is isosceles. (c) If $\angle BCP = 25°$, find $\angle PCQ$.
Record in your book -- full marks require clear working.
Reveal solution
(a) In triangles $CBP$ and $CDQ$: $CB = CD$ (sides of square), $\angle CBP = \angle CDQ = 90°$ (angles of square), $BP = DQ$ (given). Therefore $\triangle CBP \equiv \triangle CDQ$ (SAS).
(b) From congruence, $CP = CQ$ (corresponding sides). Therefore $\triangle CPQ$ is isosceles with $CP = CQ$.
(c) From congruence, $\angle DCQ = \angle BCP = 25°$. Since $\angle BCD = 90°$, we have $\angle PCQ = 90° - 25° - 25° = 40°.
R
Quick Review
SSS
Three sides equal -- rigid frame
SAS
Two sides, included angle -- hinge
AAS
Two angles, side -- shape + size
RHS
Right, hypotenuse, side -- special case
SSA fails
Ambiguous case -- two triangles possible
CPCTC
Congruent $\Rightarrow$ all parts match
Interactive: Congruence Matcher
Test your knowledge by identifying which congruence test applies to each pair of triangles. Try to decide before the interactive reveals the answer.
Consolidation Game -- Doodle Jump Quiz
+10 XP for playing
Jump your way to the top by answering questions on triangle congruence, the four tests, and formal proofs. The higher you climb, the harder the questions.
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