MathsCBSE Class 9 · Propositions and their Converses

Isosceles Triangles and the Pythagoras Converse: Class 9 Proofs

Updated 17 October 2026 · Fact-checked

In a triangle, equal sides have equal opposite angles, and (converse) equal angles have equal opposite sides. In the same way, if the squares of two sides of a triangle add up to the square of the third side then the angle opposite the third side is a right angle (converse of the Baudhāyana–Pythagoras theorem). Each converse needs its own proof; the isosceles converse uses congruence (AAS) and the Pythagoras converse is used to test right angles with side lengths.

Equal sides and equal angles imply each other, and when two squared sides add to the third squared, the angle opposite is 90°.

Illustration drawn for QuestionBuddy.

Understand Isosceles Triangles and the Converse of the Pythagoras Theorem

Every theorem has a converse, and the converse needs its own proof. Two pairs are central in Class 9 geometry.

Isosceles triangle theorem and its converse.

*Theorem:* In a triangle with $A B = A C$ , the angles opposite those sides are equal: $∠ B = ∠ C$ .
*Converse:* If $∠ B = ∠ C$ in a triangle $A B C$ , then the sides opposite are equal: $A B = A C$ .

*Proof of the converse.* Draw the bisector $A D$ of $∠ A$ meeting $B C$ at $D$ . In $△ A B D$ and $△ A C D$ : $∠ B = ∠ C$ (given), $∠ B A D = ∠ C A D$ (bisector) and $A D$ is common. By AAS the triangles are congruent, so $A B = A C$ (CPCT). Thus the isosceles theorem and its converse together say: a triangle is isosceles if and only if two of its angles are equal. If all three angles are equal ( $6 0^{\circ}$ each) then all three sides are equal. If $∠ B = ∠ C$ is given and $A B = (3 x - 4)$ , $A C = (x + 8)$ then $3 x - 4 = x + 8$ gives $x = 6$ , so $A B = A C = 14$ .

*Consequence.* In an isosceles triangle with $A B = A C$ the median $A D$ to $B C$ is also perpendicular to $B C$ : $△ A B D ≅ △ A C D$ (SSS) gives $∠ A D B = ∠ A D C$ , and these form a linear pair, so each is $9 0^{\circ}$ .

Baudhāyana–Pythagoras theorem and its converse.

*Theorem:* In a right-angled triangle, the square of the hypotenuse equals the sum of the squares of the other two sides.
*Converse:* If in a triangle the square of one side equals the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

*Using the converse.* To test whether a triangle is right-angled, compute the squares: sides $20, 21, 29$ give $400 + 441 = 841 = 2 9^{2}$ , so the angle between $20$ and $21$ is a right angle. Sides $8, 15, 18$ give $64 + 225 = 289 \neq = 324$ , so it is not right-angled. A triangle with sides $a, b, c$ must also satisfy the triangle inequality; always compare the square of the longest side with the sum of the other two squares.

*Pythagorean triples.* For any whole number $n > 1$ , the numbers $n^{2} - 1$ , $2 n$ , $n^{2} + 1$ satisfy $(n^{2} - 1)^{2} + (2 n)^{2} = n^{4} + 2 n^{2} + 1 = (n^{2} + 1)^{2}$ , so they form a right-angled triangle. $n = 6$ gives $35, 12, 37$ . Other familiar triples: $3, 4, 5$ ; $5, 12, 13$ ; $7, 24, 25$ ; $9, 12, 15$ ; $14, 48, 50$ .

*A right-angle test with rope.* A rope with knots dividing it into $3 + 4 + 5 = 12$ equal parts forms a right angle when stretched into a triangle with sides $3, 4, 5$ parts: masons used this idea to mark right angles, as in the rope constructions of the Śulbasūtras.

Which statements are reversible? The Pythagoras theorem and its converse are both true, so "a triangle is right-angled if and only if the square of the longest side equals the sum of the squares of the other two sides". On the other hand, the statement "any triangle with sides $5$ and $12$ has third side $13$ " is false: a triangle with sides $5, 12, 10$ exists and is not right-angled.

Method for proofs. Write what is given and what is to be proved; draw a construction line if needed (a bisector, a median); name the congruence rule (SAS, ASA, AAS, SSS, RHS); conclude with CPCT.

Isosceles Triangles and the Converse of the Pythagoras Theorem: formulas to remember

Isosceles theorem: $A B = A C \Rightarrow ∠ B = ∠ C$; Equal sides, equal opposite angles.
Converse of the isosceles theorem: $∠ B = ∠ C \Rightarrow A B = A C$; Proved with the angle bisector and AAS.
Baudhāyana–Pythagoras theorem: $∠ C = 9 0^{\circ} \Rightarrow a^{2} + b^{2} = c^{2}$; $c$ is the hypotenuse.
Converse: $a^{2} + b^{2} = c^{2} \Rightarrow ∠ C = 9 0^{\circ}$; Test for a right angle.
Pythagorean triple generator: $(n^{2} - 1)^{2} + (2 n)^{2} = (n^{2} + 1)^{2}$; For whole numbers $n > 1$ .
Median of an isosceles triangle: $A B = A C, B D = D C \Rightarrow A D ⊥ B C$; By SSS and a linear pair.

How to solve Isosceles Triangles and the Converse of the Pythagoras Theorem questions

Decide whether the question wants a proof or a calculation, and quote the theorem or its converse by name.

1For equal sides or angles in a triangle, decide whether you are using the theorem (sides to angles) or its converse (angles to sides).
2Set up equations from the equal sides or angles and solve for the unknown.
3For a proof of a converse, state the given and the claim, add a construction, and choose a congruence rule.
4To test a right angle, square the three sides, compare the largest square with the sum of the other two.
5Quote "converse of the Baudhāyana–Pythagoras theorem" when you conclude a right angle.
6If the converse test fails, state that the triangle is not right-angled and give the numerical comparison.

Isosceles Triangles and the Converse of the Pythagoras Theorem 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.

Converse of the Pythagoras theorem

Suppose a mason checks that the corner of a wall is square by marking 60 cm along one wall and 80 cm along the other, and then measuring the distance between the two marks, which comes out to exactly 100 cm.

Is the corner a right angle? Justify.

Try it first, then show the solutionSolution

$6 0^{2} + 8 0^{2} = 3600 + 6400 = 10000$ .
$10 0^{2} = 10000$ .
The sum of the squares of the two shorter sides equals the square of the third, so by the converse of the Baudhāyana–Pythagoras theorem the corner is a right angle.

Answer: Yes, the corner is a right angle.

Memory hook: Squares add up: the corner is square.

Tricks to remember Isosceles Triangles and the Converse of the Pythagoras Theorem

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

Equal sides, equal angles: both ways
Isosceles means two equal sides or two equal angles; each fact gives the other.
Use it when: Isosceles questions.
Square the three sides
To test a right angle, compare the largest square with the sum of the other two squares.
Use it when: Converse of Pythagoras.
Bisector plus AAS
To prove the converse of the isosceles theorem, bisect the apex angle and use AAS.
Use it when: Writing the proof.

Isosceles Triangles and the Converse of the Pythagoras Theorem: solved examples

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

Example 1: Equal angles give x · 3 marks · Short answerBoard level

In $△ A B C$ , $A B = A C$ , $∠ B = (3 x + 6)^{\circ}$ and $∠ C = (5 x - 30)^{\circ}$ . Find $x$ , all three angles, and name the type of triangle.

Try it first, then show the solutionSolution

Equal sides have equal opposite angles: $3 x + 6 = 5 x - 30$ . (1)
$2 x = 36$ , so $x = 18$ and $∠ B = ∠ C = 6 0^{\circ}$ . (1)
$∠ A = 18 0^{\circ} - 12 0^{\circ} = 6 0^{\circ}$ , so all angles are $6 0^{\circ}$ and the triangle is equilateral. (1)

Answer: $x = 18$ ; all angles $6 0^{\circ}$ ; the triangle is equilateral.

Example 2: Right-angle test · 3 marks · Short answerBoard level

Decide whether a triangle with sides (i) 20 cm, 21 cm, 29 cm and (ii) 8 cm, 15 cm, 18 cm is right-angled; if so, name the vertex with the right angle.

Try it first, then show the solutionSolution

(i) $2 0^{2} + 2 1^{2} = 400 + 441 = 841 = 2 9^{2}$ . (1)
By the converse of the Baudhāyana–Pythagoras theorem it is right-angled, with the right angle between the sides 20 and 21. (1)
(ii) $8^{2} + 1 5^{2} = 289$ but $1 8^{2} = 324$ ; they differ, so it is not right-angled. (1)

Answer: (i) right-angled (between 20 and 21); (ii) not right-angled.

Example 3: Converse of the isosceles theorem · 5 marks · Long answerHarder board level

Prove that if two angles of a triangle are equal, the sides opposite them are equal. Then in $△ P QR$ with $∠ Q = ∠ R$ , $P Q = (3 x - 4)$ and $P R = (x + 8)$ , find $x$ and $P Q$ .

Try it first, then show the solutionSolution

Given: $△ A B C$ with $∠ B = ∠ C$ . To prove: $A B = A C$ . (½)
Draw the bisector $A D$ of $∠ A$ meeting $B C$ at $D$ . (½)
In $△ A B D$ and $△ A C D$ : $∠ B = ∠ C$ (given), $∠ B A D = ∠ C A D$ (bisector), $A D = A D$ (common). (1)
So $△ A B D ≅ △ A C D$ (AAS), hence $A B = A C$ (CPCT). (1)
Application: $P Q$ is opposite $∠ R$ and $P R$ is opposite $∠ Q$ ; with $∠ Q = ∠ R$ , $P Q = P R$ , so $3 x - 4 = x + 8$ , $x = 6$ . (1)
$P Q = 3 (6) - 4 = 14$ ; check $P R = 6 + 8 = 14$ . (1)

Answer: The converse is proved; $x = 6$ and $P Q = P R = 14$ .

Example 4: Generating right triangles · 3 marks · Short answerHarder board level

Show that for every whole number $n > 1$ the numbers $n^{2} - 1$ , $2 n$ and $n^{2} + 1$ are the sides of a right-angled triangle, and find the triple for $n = 6$ .

Try it first, then show the solutionSolution

$(n^{2} - 1)^{2} + (2 n)^{2} = n^{4} - 2 n^{2} + 1 + 4 n^{2} = n^{4} + 2 n^{2} + 1$ . (1)
$= (n^{2} + 1)^{2}$ , so by the converse of the Pythagoras theorem the triangle is right-angled. (1)
For $n = 6$ : $35, 12, 37$ ; check $3 5^{2} + 1 2^{2} = 1225 + 144 = 1369 = 3 7^{2}$ . (1)

Answer: The triple is $(35, 12, 37)$ for $n = 6$ .

Example 5: Is the frame square? · 4 marks · Case-basedHOTS / competency

Case: a carpenter builds a rectangular frame with sides 15 cm and 8 cm. (i) What diagonal should it have if the corners are right angles? (ii) She measures the diagonal as 18 cm: is the corner a right angle? (iii) A second triangular brace has sides 9 cm, 12 cm and 15 cm: is it right-angled? (iv) Write the converse of the theorem you used and say whether it is true.

Try it first, then show the solutionSolution

(i) Diagonal $= 1 5^{2} + 8^{2} = 289 = 17$ cm. (1)
(ii) With the measured 18 cm: $1 5^{2} + 8^{2} = 289 \neq = 324 = 1 8^{2}$ , so the corner is not a right angle. (1)
(iii) $9^{2} + 1 2^{2} = 81 + 144 = 225 = 1 5^{2}$ , so the brace is right-angled. (1)
(iv) Theorem: in a right triangle, $a^{2} + b^{2} = c^{2}$ . Converse: if $a^{2} + b^{2} = c^{2}$ in a triangle, then the angle opposite $c$ is a right angle. It is true. (1)

Answer: (i) 17 cm; (ii) no; (iii) yes; (iv) the converse is true.

Where students lose marks in Isosceles Triangles and the Converse of the Pythagoras Theorem

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

Mistake: Using the theorem when the converse is needed
Assuming a right angle to prove it.
Wrong working
"Since $a^{2} + b^{2} = c^{2}$ by Pythagoras, the angle is $9 0^{\circ}$ ."
Right working
By the converse of the Pythagoras theorem, $a^{2} + b^{2} = c^{2}$ implies the angle is $9 0^{\circ}$ .
Do this: Name the converse when going from squares to the angle.
Usually costs: 1 mark.
Mistake: Testing the wrong side as the hypotenuse
Comparing a smaller square with the sum.
Wrong working
$8^{2} + 1 8^{2} \neq = 1 5^{2}$ so not right-angled.
Right working
Test the largest side: $8^{2} + 1 5^{2} = 289 \neq = 1 8^{2} = 324$ .
Do this: Square the longest side first.
Usually costs: 1 to 2 marks.
Mistake: Pairing sides with the wrong angles
Equating adjacent angle and side.
Wrong working
$∠ B = ∠ C$ so $A B = B C$ .
Right working
The sides opposite $∠ B$ and $∠ C$ are $A C$ and $A B$ , so $A B = A C$ .
Do this: Write "opposite" explicitly.
Usually costs: The whole answer.

Exam tips

Name the rule (AAS, SSS, CPCT) at each step of a proof.
Write "sides opposite equal angles" in full when using the converse.
Test the largest side for the right-angle check.
State the final conclusion in words as well as symbols.

Practice questions on Isosceles Triangles and the Converse of the Pythagoras Theorem

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 can be the sides of a right-angled triangle?
(a) 6, 8, 11
(b) 7, 24, 25
(c) 5, 9, 12
(d) 9, 10, 13
Hint
Square the three numbers and test the largest.
Answer
(b). $7^{2} + 2 4^{2} = 49 + 576 = 625 = 2 5^{2}$ .
Q2 · 3 marks · Short answerBoard level
In $△ A B C$ , $∠ B = ∠ C$ , $A B = (2 y + 1)$ cm, $A C = (3 y - 5)$ cm and $B C = 10$ cm. Find $y$ and the perimeter.
Hint
Equal angles mean equal opposite sides: $A B = A C$ .
Answer
$A B = A C$ (sides opposite equal angles), so $2 y + 1 = 3 y - 5$ and $y = 6$ . Then $A B = A C = 13$ cm and the perimeter $= 13 + 13 + 10 = 36$ cm.
Q3 · 3 marks · Short answerHarder board level
In $△ A B C$ with $A B = A C$ , $D$ is the midpoint of $B C$ . Prove that $A D ⊥ B C$ .
Hint
Use SSS on $△ A B D$ and $△ A C D$ .
Answer
In $△ A B D$ and $△ A C D$ : $A B = A C$ (given), $B D = D C$ ( $D$ is the midpoint), $A D = A D$ (common). So the triangles are congruent (SSS) and $∠ A D B = ∠ A D C$ (CPCT). These two angles form a linear pair, so their sum is $18 0^{\circ}$ and each is $9 0^{\circ}$ . Hence $A D ⊥ B C$ .
Q4 · 5 marks · Long answerHarder board level
In quadrilateral $A B C D$ , $A B = 9$ cm, $B C = 12$ cm, $C D = 8$ cm, $D A = 17$ cm and the diagonal $A C = 15$ cm. Show that $∠ A B C = 9 0^{\circ}$ and $∠ A C D = 9 0^{\circ}$ , and find the area of the quadrilateral.
Hint
Test each triangle with the converse of the Pythagoras theorem.
Answer
In $△ A B C$ : $9^{2} + 1 2^{2} = 81 + 144 = 225 = 1 5^{2}$ , so $∠ A B C = 9 0^{\circ}$ . In $△ A C D$ : $1 5^{2} + 8^{2} = 225 + 64 = 289 = 1 7^{2}$ , so $∠ A C D = 9 0^{\circ}$ . Areas: $\frac{1}{2} \times 9 \times 12 = 54$ and $\frac{1}{2} \times 15 \times 8 = 60$ . Total area $= 114$ cm $^{2}$ .
Q5 · 4 marks · Case-basedHOTS / competency
Case: a student claims "Any triangle with two sides of length 5 and 12 must have its third side 13." (i) State the true statement about a right-angled triangle with legs 5 and 12. (ii) Is the claim the converse of that statement? (iii) Give a counterexample to the claim using a valid triangle. (iv) Find the third side if 12 is the hypotenuse and 5 is a leg.
Hint
The third side can be any length between 7 and 17 (triangle inequality).
Answer
(i) If a triangle has a right angle between sides 5 and 12, the third side is 13. (ii) No; the claim says the third side must be 13 for any triangle with sides 5 and 12, which is a different statement. (iii) A triangle with sides $5, 12, 10$ exists ( $5 + 10 > 12$ ) and the third side is 10, not 13; it is not right-angled since $25 + 100 = 125 \neq = 144$ . (iv) $1 2^{2} - 5^{2} = 119$ .

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.Converse of the isosceles triangle theorem?Show answer
Sides opposite equal angles are equal.
2.Which congruence rule proves it?Show answer
AAS, after drawing the angle bisector.
3.Converse of the Pythagoras theorem?Show answer
If $a^{2} + b^{2} = c^{2}$ , the angle opposite $c$ is $9 0^{\circ}$ .
4.A triple from $n = 5$ ?Show answer
$(24, 10, 26)$ .
5.Can a triangle with sides 5, 12 and 10 be right-angled?Show answer
No, $25 + 100 \neq = 144$ .

Isosceles Triangles and the Converse of the Pythagoras Theorem: frequently asked questions

Why must I prove the converse separately?

Because a theorem being true does not make its converse true; each statement is a separate claim, and each needs its own proof or a counterexample.

Is every isosceles triangle a right triangle?

No. Only the isosceles triangle with apex angle 90^\circ (equal sides 1, 1, hypotenuse \sqrt{2}, for example) is right-angled.

What is a Pythagorean triple?

Three whole numbers a, b, c with a^2 + b^2 = c^2, such as 3, 4, 5.