Properties of triangles

Two sides of a triangle are equal — what does that force on the angles? The Class 9 chapter on triangles answers this with one clean, provable fact: the angles facing equal sides are themselves equal, and the reverse holds too. That single idea, plus a companion pair of "no-proof-needed" inequalities about how a triangle's three sides and angles constrain each other, is what "properties of triangles" means in this chapter — as distinct from the earlier chapter on proving two whole triangles congruent. Here the spotlight turns inward, onto the relationships that live inside one triangle by itself. The isosceles triangle theorem is the centrepiece, built from a congruence proof you already know (SAS), and it unlocks a shortcut used constantly in geometry: spot equal sides, get equal angles for free, or spot equal angles and get equal sides for free. The triangle inequality and the angle-side relation round out the toolkit, letting you check whether three lengths can even form a triangle, and compare angles or sides without a protractor or ruler.

The core insight

In any triangle, angles opposite equal sides are equal (and conversely), while the sum of any two sides must exceed the third, and a larger angle always sits opposite a longer side.

The core idea

Isosceles triangle theorem. If two sides of a triangle are equal, the angles opposite those sides are equal. In $\triangle ABC$ , if $AB = AC$ then:

\angle B = \angle C

where: $\angle B$ is the angle at vertex $B$ , opposite side $AC$ , and $\angle C$ is the angle at vertex $C$ , opposite side $AB$ . This is proved in Class 9 using the SAS congruence criterion — see the derivation below.

Converse (stated without proof). Run the theorem backwards: if two angles of a triangle are equal, the sides opposite those angles are equal. So $\angle B = \angle C$ in $\triangle ABC$ gives $AB = AC$ . Combined, the theorem and its converse mean a triangle is isosceles exactly when two of its angles are equal — neither direction needs re-deriving once you know the other holds. A quick corollary: in an equilateral triangle all three sides are equal, so applying the theorem pairwise to each side shows all three angles are equal too. Since the three angles of any triangle sum to $180°$ , each angle of an equilateral triangle is $60°$ .

Triangle inequality (stated without proof). The sum of any two sides of a triangle must be greater than the third side:

AB + BC > CA, \qquad BC + CA > AB, \qquad CA + AB > BC

where: $AB$ , $BC$ , $CA$ are the three side lengths of $\triangle ABC$ . In practice, only the smallest two sides need checking against the largest — if their sum beats the largest side, all three inequalities automatically hold.

Angle-side relation (stated without proof). The side opposite a larger angle is longer, and conversely the angle opposite a longer side is larger. If $\angle B > \angle C$ in $\triangle ABC$ , then $AC > AB$ . This is a strict inequality version of the isosceles theorem: equal angles give equal sides, and now unequal angles give sides unequal in the same order.

Figure

An isosceles triangle with the two equal sides marked by tick marks and the two base angles marked equal, alongside the bisector from the apex meeting the base at a right angle

Derivation AProving the isosceles triangle theorem

The goal is to show, using nothing beyond the SAS congruence criterion already available from the congruence chapter, that if two sides of a triangle are equal then the two angles opposite those sides must also be equal — the reasoning that justifies why the isosceles theorem is a proved result in Class 9 rather than merely an observed pattern.

Step 1. Set up the triangle and the construction. Let $\triangle ABC$ have $AB = AC$ . Draw the bisector of $\angle A$ , and let it meet $BC$ at point $D$ , so that $\angle BAD = \angle CAD$ . This construction splits the original triangle into two smaller triangles, $\triangle ABD$ and $\triangle ACD$ , sharing the side $AD$ .

Step 2. Compare $\triangle ABD$ and $\triangle ACD$ pair by pair. $AB = AC$ by the given condition. $\angle BAD = \angle CAD$ because $AD$ bisects $\angle A$ by construction. $AD = AD$ since it is the common side to both triangles. These are exactly two sides and the included angle of one triangle matching two sides and the included angle of the other.

Step 3. Apply SAS. Two sides and the included angle of $\triangle ABD$ equal the corresponding two sides and included angle of $\triangle ACD$ , so by the SAS congruence criterion $\triangle ABD \cong \triangle ACD$ .

Step 4. Read off the result by CPCT. Corresponding parts of congruent triangles are equal, so $\angle ABD = \angle ACD$ , which is exactly $\angle B = \angle C$ . The same congruence also gives $BD = DC$ and $\angle ADB = \angle ADC$ ; since these two angles lie on the straight line $BC$ and are equal, each must be $90°$ . So the bisector of the apex angle of an isosceles triangle also bisects the base at right angles — a free bonus from the same proof.

Worked example

In $\triangle PQR$ , $PQ = PR$ and $\angle Q = 70°$ . Find $\angle P$ and $\angle R$ , and check whether a triangle with sides $5\text{ cm}$ , $6\text{ cm}$ , $12\text{ cm}$ could exist instead.

Step 1. Identify the equal sides and apply the isosceles theorem. Since $PQ = PR$ , the angles opposite these sides are equal: the angle opposite $PQ$ is $\angle R$ , and the angle opposite $PR$ is $\angle Q$ . So $\angle R = \angle Q = 70°$ .

Step 2. Use the angle sum property to find the remaining angle. $\angle P + \angle Q + \angle R = 180°$ , so $\angle P + 70° + 70° = 180°$ , giving $\angle P = 180° - 140° = 40°$ .

Step 3. Now test the second, unrelated set of lengths against the triangle inequality. For $5\text{ cm}$ , $6\text{ cm}$ , $12\text{ cm}$ , check the two smaller sides against the largest: $5 + 6 = 11$ , which is not greater than $12$ . The inequality fails, so no such triangle exists — three lengths only form a triangle when the two smallest together outreach the largest.

Answer: $\angle P = 40°$ , $\angle R = 70°$ , and the $5\text{ cm}, 6\text{ cm}, 12\text{ cm}$ triangle is impossible.

Competitive stretch

For sides $7\text{ cm}$ , $10\text{ cm}$ , $13\text{ cm}$ , first confirm existence: $7 + 10 = 17 > 13$ , so the triangle exists. Then, without measuring any angle, the angle-side relation says the largest angle sits opposite the longest side — here $13\text{ cm}$ — so that angle is the largest in the triangle, decided purely from the three lengths.

Common traps

Mixing up which side is "opposite" which angle — in $\triangle ABC$ , $\angle A$ sits opposite side $BC$ , not side $AB$ or $AC$ ; get this backwards and the isosceles theorem gets applied to the wrong pair.
Checking the triangle inequality against only one pair of sides and stopping there, instead of checking (or at least recognising it suffices to check) the two smallest sides against the largest.
Treating the converse of the isosceles theorem as needing a fresh SAS-style proof — it is a stated, without-proof result in Class 9 and can simply be quoted.
Swapping the direction of the angle-side relation — "larger angle gives longer opposite side" and "longer side gives larger opposite angle" are the same fact read two ways, but it is easy to accidentally apply it to an adjacent side or angle instead of the opposite one.

Ask the tutor: if you are unsure which angle is opposite a given side in a labelled figure, or want your triangle-inequality check walked through step by step, ask the AI tutor to work the diagram with you.

Why is the isosceles theorem proved but the triangle inequality just stated without proof in Class 9

The isosceles theorem has a short, accessible proof using only the SAS congruence criterion students already know, so the syllabus asks for it in full. The triangle inequality and the angle-side relation, by contrast, need proof techniques (extending a side, comparing exterior angles) that go slightly beyond what has been built up at this stage, so Class 9 introduces them as tools to use rather than results to derive. You are expected to apply both confidently, just not reproduce a proof for them on demand.

Does the isosceles theorem work if I only know two angles are equal, not two sides

Yes, but that is a different (though related) statement — the converse. If $\angle B = \angle C$ in a triangle, then $AB = AC$ , and you can quote this directly without proving it fresh; it does not need the bisector construction used for the original theorem. The two statements together mean equal sides and equal angles always travel in pairs in a triangle, so spotting either one instantly hands you the other.

How is the angle-side relation different from the isosceles theorem

The isosceles theorem and its converse handle equality: equal sides go with equal angles, and vice versa. The angle-side relation extends this to inequality: when the sides are not equal, the larger angle sits opposite the longer side, and the larger side sits opposite the larger angle. Think of it as the same underlying relationship stretched from an "if equal, then equal" statement into an "if bigger, then bigger" statement — useful whenever you need to compare, not just match, sides and angles.

Properties of triangles

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads