MathsCBSE Class 9 · Triangles: Congruence Theorems

SSS and RHS Congruence, Equidistant Points and Tests That Fail

Updated 17 October 2026 · Fact-checked

Two triangles are congruent by SSS if their three sides are equal in pairs, and by RHS if they are right-angled with equal hypotenuses and one other pair of equal sides. AAA (only angles) and SSA (two sides and a non-included angle) do not prove congruence. Congruence also proves that points on the perpendicular bisector of a segment are equidistant from its ends, and points on an angle bisector are equidistant from the arms.

Understand SSS and RHS Congruence, Equidistant Points and Failed Tests

SSS rule. If the three sides of one triangle are equal to the three sides of another, the triangles are congruent. A triangle with sides $30, 40, 50$ cut from three rods can only be built one way, so two such braces are identical.

RHS rule. If two right-angled triangles have equal hypotenuses and one pair of equal sides (legs), they are congruent (Right angle, Hypotenuse, Side). The other leg then follows from Pythagoras: hypotenuse $13$ and leg $5$ give the other leg $169 - 25 = 12$ in both triangles. This is a special case of "SSA" that does work, because the right angle removes the ambiguity.

Why AAA fails. Equal angles give triangles of the same *shape* but not necessarily the same size: the triangles with sides $3, 4, 5$ and $6, 8, 10$ both have angles about $3 7^{\circ}$ , $5 3^{\circ}$ and $9 0^{\circ}$ but are not congruent.

Why SSA fails. Take an isosceles triangle $A B C$ with $A B = A C$ and a point $D$ on $B C$ that is not the midpoint. In $△ A B D$ and $△ A C D$ we have $A B = A C$ , $A D = A D$ (common) and $∠ B = ∠ C$ , an angle opposite the common side $A D$ in each triangle. This is the SSA pattern, yet $B D \neq = D C$ , so the triangles are not congruent. Hence two sides and a non-included angle are not enough (except in the right-angled case, RHS).

Equidistance on the perpendicular bisector. *Theorem:* a point $P$ on the perpendicular bisector of segment $A B$ is equidistant from $A$ and $B$ . *Proof:* let $M$ be the midpoint of $A B$ . In $△ P M A$ and $△ P M B$ : $P M = P M$ (common), $∠ P M A = ∠ P M B = 9 0^{\circ}$ , $A M = B M$ ; so they are congruent by SAS and $P A = P B$ (CPCT). *Converse:* if $P A = P B$ , then $P$ lies on the perpendicular bisector of $A B$ . *Proof:* with $M$ the midpoint, $△ P M A$ and $△ P M B$ have $P A = P B$ , $A M = B M$ and $P M$ common, so they are congruent (SSS); hence $∠ P M A = ∠ P M B$ , and these form a linear pair, so each is $9 0^{\circ}$ and $P M$ is the perpendicular bisector.

Equidistance on the angle bisector. *Theorem:* a point $P$ on the bisector of $∠ A O B$ is equidistant from the arms $O A$ and $O B$ . *Proof:* draw $P M ⊥ O A$ and $P N ⊥ O B$ . In $△ O P M$ and $△ O P N$ : $∠ P M O = ∠ P N O = 9 0^{\circ}$ , $∠ P O M = ∠ P O N$ (bisector), $O P = O P$ ; so they are congruent (AAS) and $P M = P N$ . If $P M = (2 x + 3)$ and $P N = (5 x - 6)$ then $x = 3$ and the distance is $9$ .

The kite. In a quadrilateral $A B C D$ with $A B = A D$ and $C B = C D$ , the diagonal $A C$ is common, so $△ A B C ≅ △ A D C$ (SSS). Hence $A C$ bisects $∠ A$ and $∠ C$ , and $A C$ is the perpendicular bisector of $B D$ . For $A B = A D = 10$ , $C B = C D = 17$ and $B D = 16$ , with $O$ the midpoint of $B D$ : $B O = 8$ , $A O = 100 - 64 = 6$ , $O C = 289 - 64 = 15$ , so $A C = 21$ and the area is $\frac{1}{2} \times 21 \times 16 = 168$ .

Method. Decide if sides alone (SSS), a right angle with hypotenuse and a side (RHS), or a construction (midpoint, bisector, perpendicular) is available; list the three pairs with reasons; conclude with CPCT.

SSS and RHS Congruence, Equidistant Points and Failed Tests: formulas to remember

SSS: $A B = P Q, B C = QR, C A = R P$; All three sides.
RHS: $∠ B = ∠ Q = 9 0^{\circ}, A C = P R, A B = P Q$; Hypotenuse and one leg.
Not congruence rules: $AAA, SSA (except RHS)$; Same shape or ambiguous.
Perpendicular bisector: $P on the perpendicular bisector of A B ⟺ P A = P B$; Both directions.
Angle bisector: $P on the bisector of ∠ A O B \Rightarrow P M = P N$; Perpendicular distances to the arms.
Kite: $A B = A D, C B = C D \Rightarrow A C ⊥ B D, A C bisects B D$; By SSS then SAS.

How to solve SSS and RHS Congruence, Equidistant Points and Failed Tests questions

Pick SSS or RHS when sides are given, or build a construction to create them.

1List all given equal sides and mark right angles and common sides.
2If three pairs of sides can be matched, use SSS; if the triangles are right-angled, look for the hypotenuse and one side for RHS.
3For points equidistant from two others, join to the midpoint (or draw perpendiculars to the arms) and compare two triangles.
4Write the three pairs in order, each with a reason, then the rule.
5Apply CPCT to get angles or sides.
6For numerical parts use Pythagoras in the right triangles that appear.

SSS and RHS Congruence, Equidistant Points and Failed Tests in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

SSS congruence

Suppose a carpenter cuts two braces for a gate, each made of rods 30 cm, 40 cm and 50 cm joined end to end.

Are the two triangular braces identical? Is the corner between the 30 cm and 40 cm rods a right angle?

Try it first, then show the solutionSolution

All three pairs of sides are equal, so the braces are congruent (SSS).
$3 0^{2} + 4 0^{2} = 900 + 1600 = 2500 = 5 0^{2}$ , so by the converse of the Baudhāyana–Pythagoras theorem the angle between the 30 cm and 40 cm rods is a right angle in both braces.

Answer: Yes, the braces are congruent by SSS, and the corner between the 30 cm and 40 cm rods is a right angle.

Memory hook: Three sides lock the shape.

Tricks to remember SSS and RHS Congruence, Equidistant Points and Failed Tests

Short memory aids. Say each one aloud once, then use it on a question the same day.

Right, hypotenuse, side
RHS needs a right angle in both, equal hypotenuses and one equal leg.
Use it when: Right-angled triangle proofs.
AAA same shape, SSA ambiguous
Equal angles fix the shape, not the size; two sides and a non-included angle can fit two ways.
Use it when: Remembering what is not a rule.
Join to the midpoint
To prove a point is equidistant from two ends, join it to the midpoint and compare the two triangles.
Use it when: Perpendicular bisector proofs.

SSS and RHS Congruence, Equidistant Points and Failed Tests: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Kite and SSS · 3 marks · Short answerBoard level

In quadrilateral $A B C D$ , $A B = A D$ and $C B = C D$ . Prove that $△ A B C ≅ △ A D C$ , that $∠ B = ∠ D$ and that $A C$ bisects $∠ B A D$ .

Try it first, then show the solutionSolution

In $△ A B C$ and $△ A D C$ : $A B = A D$ , $C B = C D$ , $A C = A C$ (common). (1)
So $△ A B C ≅ △ A D C$ (SSS). (1)
By CPCT $∠ B = ∠ D$ and $∠ B A C = ∠ D A C$ , so $A C$ bisects $∠ B A D$ . (1)

Answer: The triangles are congruent by SSS; $∠ B = ∠ D$ and $A C$ bisects $∠ B A D$ .

Example 2: RHS · 3 marks · Short answerBoard level

$△ A B C$ is right-angled at $B$ and $△ D E F$ is right-angled at $E$ , with $A C = D F = 13$ cm and $A B = D E = 5$ cm. Prove that they are congruent, and find $B C$ , $E F$ and the area.

Try it first, then show the solutionSolution

$∠ B = ∠ E = 9 0^{\circ}$ , hypotenuse $A C = D F$ , side $A B = D E$ : $△ A B C ≅ △ D E F$ (RHS). (1½)
$B C = 1 3^{2} - 5^{2} = 144 = 12$ cm, and $E F = B C = 12$ cm (CPCT). (1)
Area $= \frac{1}{2} \times 5 \times 12 = 30$ cm $^{2}$ . (½)

Answer: Congruent by RHS; $B C = E F = 12$ cm; area $30$ cm $^{2}$ .

Example 3: Perpendicular bisector and its converse · 5 marks · Long answerHarder board level

Prove that a point on the perpendicular bisector of a segment $A B$ is equidistant from $A$ and $B$ , and that a point equidistant from $A$ and $B$ lies on its perpendicular bisector.

Try it first, then show the solutionSolution

Let $M$ be the midpoint of $A B$ and $P$ a point on the perpendicular to $A B$ at $M$ . In $△ P M A$ and $△ P M B$ : $A M = B M$ , $∠ P M A = ∠ P M B = 9 0^{\circ}$ , $P M = P M$ . (1½)
By SAS $△ P M A ≅ △ P M B$ , so $P A = P B$ (CPCT). (1)
Converse: let $P A = P B$ and join $P$ to the midpoint $M$ of $A B$ . (½)
In $△ P M A$ and $△ P M B$ : $P A = P B$ , $A M = B M$ , $P M = P M$ ; so they are congruent (SSS). (1)
$∠ P M A = ∠ P M B$ , and as a linear pair they add to $18 0^{\circ}$ , so each is $9 0^{\circ}$ : $P M$ is the perpendicular bisector. (1)

Answer: Both statements are proved.

Example 4: Tests that fail · 3 marks · Short answerHarder board level

Give an example to show that (i) AAA and (ii) SSA do not prove two triangles congruent.

Try it first, then show the solutionSolution

(i) The triangles with sides $3, 4, 5$ and $6, 8, 10$ both have angles about $3 7^{\circ}$ , $5 3^{\circ}$ , $9 0^{\circ}$ , but one is twice as large: not congruent. (1½)
(ii) In isosceles $△ A B C$ with $A B = A C$ , take $D$ on $B C$ which is not the midpoint. In $△ A B D$ and $△ A C D$ : $A B = A C$ , $A D = A D$ , $∠ B = ∠ C$ (SSA pattern), but $B D \neq = D C$ , so they are not congruent. (1½)

Answer: AAA is shown by $3, 4, 5$ and $6, 8, 10$ ; SSA by $△ A B D$ and $△ A C D$ above.

Example 5: A kite-shaped plot · 4 marks · Case-basedHOTS / competency

Case: a plot $A B C D$ is a kite with $A B = A D = 10$ m, $C B = C D = 17$ m and the diagonal $B D = 16$ m, meeting the diagonal $A C$ at $O$ . (i) Prove $△ A B C ≅ △ A D C$ . (ii) Prove that $A C ⊥ B D$ and $B O = O D$ . (iii) Find $A O$ , $O C$ and $A C$ . (iv) Find the area of the plot.

Try it first, then show the solutionSolution

(i) $A B = A D$ , $C B = C D$ , $A C$ common: SSS. (½)
(ii) $∠ B A C = ∠ D A C$ (CPCT), so in $△ A B O$ and $△ A D O$ : $A B = A D$ , $∠ B A O = ∠ D A O$ , $A O$ common (SAS); hence $B O = O D$ and $∠ A O B = ∠ A O D = 9 0^{\circ}$ (linear pair). (1½)
(iii) $B O = 8$ ; $A O = 1 0^{2} - 8^{2} = 6$ ; $O C = 1 7^{2} - 8^{2} = 15$ ; $A C = 21$ m. (1)
(iv) Area $= \frac{1}{2} \times A C \times B D = \frac{1}{2} \times 21 \times 16 = 168$ m $^{2}$ . (1)

Answer: (i) and (ii) proved; $A O = 6$ , $O C = 15$ , $A C = 21$ m; area $168$ m $^{2}$ .

Where students lose marks in SSS and RHS Congruence, Equidistant Points and Failed Tests

The slips examiners see most often, and what to write instead.

Mistake: Using SSA as a rule
Two sides and any angle seem enough.
Wrong working
$A B = D E$ , $B C = E F$ , $∠ A = ∠ D$ , so congruent by SAS.
Right working
The angle is not included; SSA does not prove congruence (except RHS).
Do this: Check the position of the angle.
Usually costs: The proof.
Mistake: Using RHS without the right angle
Forgetting to state the right angles.
Wrong working
Hypotenuse and a side equal, so RHS.
Right working
State " $∠ B = ∠ E = 9 0^{\circ}$ " first.
Do this: Name the right angles in the proof.
Usually costs: 1 mark.
Mistake: Concluding perpendicular without the linear pair
Stopping at equal angles.
Wrong working
$∠ P M A = ∠ P M B$ , so $P M ⊥ A B$ .
Right working
$∠ P M A = ∠ P M B$ and they form a linear pair, so each is $9 0^{\circ}$ .
Do this: Add the linear pair reason.
Usually costs: 1 mark.

Exam tips

Write the right-angle statement before using RHS.
Use a midpoint or perpendicular construction when the figure gives no triangles to compare.
Never quote AAA or SSA as a congruence rule.
State both directions when a question asks for a theorem and its converse.

Practice questions on SSS and RHS Congruence, Equidistant Points and Failed Tests

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Which of the following is NOT a rule for proving two triangles congruent?
(a) SAS
(b) SSA
(c) RHS
(d) AAS
Hint
One pattern fails except in right-angled triangles.
Answer
(b) SSA. Two sides and a non-included angle do not fix the triangle (except in the right-angled case, RHS).
Q2 · 3 marks · Short answerBoard level
In right triangles $P QR$ (right angle at $Q$ ) and $X Y Z$ (right angle at $Y$ ), $P R = X Z = 17$ cm and $QR = Y Z = 8$ cm. Prove congruence, state the rule, and find $P Q$ .
Hint
The hypotenuses and one leg are equal.
Answer
Right angles at $Q$ and $Y$ , hypotenuses $P R = X Z$ , legs $QR = Y Z$ : $△ P QR ≅ △ X Y Z$ (RHS). $P Q = 1 7^{2} - 8^{2} = 225 = 15$ cm, and $X Y = 15$ cm (CPCT).
Q3 · 3 marks · Short answerHarder board level
$△ A B C$ and $△ D C B$ are on the same base $B C$ , on the same side, with $A B = D C$ and $A C = D B$ . Prove that $∠ A B C = ∠ D C B$ .
Hint
$B C$ is common to both triangles.
Answer
In $△ A B C$ and $△ D C B$ : $A B = D C$ , $A C = D B$ , $B C = C B$ (common). So $△ A B C ≅ △ D C B$ (SSS), and $∠ A B C = ∠ D C B$ (CPCT).
Q4 · 5 marks · Long answerHarder board level
$P$ is a point on the bisector of $∠ A O B$ ; $P M ⊥ O A$ and $P N ⊥ O B$ . Prove that $P M = P N$ . If $P M = (2 x + 3)$ cm and $P N = (5 x - 6)$ cm, find $x$ and the distance of $P$ from each arm.
Hint
Use AAS on $△ O P M$ and $△ O P N$ .
Answer
In $△ O P M$ and $△ O P N$ : $∠ P M O = ∠ P N O = 9 0^{\circ}$ , $∠ P O M = ∠ P O N$ ( $O P$ bisects $∠ A O B$ ), $O P = O P$ (common). So the triangles are congruent (AAS) and $P M = P N$ (CPCT). Then $2 x + 3 = 5 x - 6$ gives $x = 3$ , and $P M = P N = 9$ cm.
Q5 · 4 marks · Case-basedHOTS / competency
Case: two ladders lean against a vertical wall on level ground. Each ladder is 13 m long and each reaches a height of 12 m on the wall. (i) Are the triangles formed by the ladders, the wall and the ground congruent, and by which rule? (ii) How far is the foot of each ladder from the wall? (iii) What would be missing if only the ladder lengths (13 m) and the ground distance (5 m) were known and the ground and wall were not known to be at right angles? (iv) Why is AAA of no use here?
Hint
The wall is vertical, so the angle with the ground is $9 0^{\circ}$ .
Answer
(i) Both are right-angled, with equal hypotenuses (13 m) and an equal leg (12 m): congruent by RHS. (ii) $1 3^{2} - 1 2^{2} = 25 = 5$ m. (iii) The right angle between wall and ground is missing. Two sides (13 m and 5 m) with no known angle, or with only a non-included angle, is not enough, since SSA does not prove congruence. RHS needs the right angle stated, and then the wall height follows by Pythagoras as 12 m. (iv) Equal angles would only show the triangles are similar in shape, not the same size: a shorter ladder could make the same angles.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.What does RHS stand for?Show answer
Right angle, hypotenuse, side.
2.Is AAA a congruence rule?Show answer
No, it only gives the same shape.
3.A point on the perpendicular bisector of AB?Show answer
It is equidistant from A and B.
4.A point on the bisector of an angle?Show answer
It is equidistant from the two arms.
5.Rule for the kite diagonal split?Show answer
SSS.

SSS and RHS Congruence, Equidistant Points and Failed Tests: frequently asked questions

Why does RHS work although SSA does not?

The right angle makes the third side determined by Pythagoras, so there is no second possible triangle.

Can two triangles with the same angles be congruent?

Yes, if they are also the same size; equal angles alone are not enough.

When do I draw a perpendicular?

When the question mentions distance from a point to a line or an arm, since distance means the perpendicular distance.