SAS Similarity Theorem (Theorem 6.5) Class 10

The SAS Similarity Theorem (Theorem 6.5) is one of the similarity criteria in Class 10 Maths chapter 6 (Triangles): NCERT prints the statement on page 90 and proves it on pages 90–91. This page gives the book’s exact statement, its proof figure (Fig 6.28), and a step-by-step proof that answers the two guided prompts the book leaves for you.

The statement of the SAS similarity theorem, exactly as NCERT prints it

Theorem 6.5 appears on printed page 90 of the NCERT Class 10 Mathematics textbook, in chapter 6 (Triangles). This is the sentence to reproduce in the exam, word for word:

Theorem 6.5 (NCERT Class 10 Mathematics, page 90) If one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

What the SAS similarity criterion means, in plain words

Take two triangles. If one angle of the first triangle equals one angle of the second, and the two sides that form that angle are in the same ratio in both triangles, then the triangles are similar — that is this theorem (NCERT, p. 90).

The load-bearing word is included. Angle A sits between sides AB and AC; angle D sits between sides DE and DF. The two sides you put into the ratio must be the two sides that enclose the equal angle, not any other pair.

Before stating the theorem, the book runs a motivating activity (NCERT, p. 90) with Fig 6.27: two triangles with \(\frac{AB}{DE} = \frac{AC}{DF}\) (each equal to \(\frac{2}{3}\)) and \(\angle A = \angle D\).

Measuring \(\angle B\), \(\angle C\), \(\angle E\) and \(\angle F\) gives \(\angle B = \angle E\) and \(\angle C = \angle F\), so the triangles are similar by the AAA criterion. Theorem 6.5 turns that single observation into a general result.

The figure NCERT uses for the theorem: Fig 6.28

Fig 6.28 (NCERT, p. 90) is the proof figure. It shows triangle ABC and a larger triangle DEF, chosen so that \(\frac{AB}{DE} = \frac{AC}{DF}\) and \(\angle A = \angle D\). The ratio is taken less than 1, which means the sides of ABC are shorter than the matching sides of DEF — so ABC fits inside DEF.

That is why the construction works. A point P is cut on side DE so that DP = AB, and a point Q is cut on side DF so that DQ = AC, then PQ is joined. Triangle DPQ is an exact copy of ABC sitting inside DEF; the rest of the proof shows this copy lines up with DEF.

Given, to prove and construction

• Given: in triangles ABC and DEF, \(\frac{AB}{DE} = \frac{AC}{DF}\) \((\lt 1)\) and \(\angle A = \angle D\).
• To prove: \(\Delta ABC \sim \Delta DEF\).
• Construction: cut \(DP = AB\) on side DE and \(DQ = AC\) on side DF, then join PQ.

This construction is the whole trick of the proof: the cut points place a copy of ABC inside DEF, so the remaining work is just comparing angles.

Proof of Theorem 6.5, step by step

The book’s proof is guided: it states two key facts and asks (How?) why they hold, then asks (Why?) the final similarity follows. Answering those prompts in order gives a complete exam-ready proof.

1. Set up the given facts and the construction. We have \(\frac{AB}{DE} = \frac{AC}{DF}\) \((\lt 1)\) and \(\angle A = \angle D\). We cut \(DP = AB\) on DE, \(DQ = AC\) on DF, and join PQ.
2. Show \(\Delta ABC \cong \Delta DPQ\). How? — By construction \(DP = AB\) and \(DQ = AC\), and the angles between these pairs are equal because \(\angle A = \angle D\). The SAS congruence rule therefore gives \(\Delta ABC \cong \Delta DPQ\), and congruence gives \(\angle B = \angle P\) and \(\angle C = \angle Q\).
3. Show \(PQ \parallel EF\). How? — Since \(DP = AB\) and \(DQ = AC\), the given ratio \(\frac{AB}{DE} = \frac{AC}{DF}\) becomes \(\frac{DP}{DE} = \frac{DQ}{DF}\). Points P and Q divide the two sides of triangle DEF in the same ratio, so by the converse of the basic proportionality theorem, \(PQ \parallel EF\).
4. Transfer the angles. Because \(PQ \parallel EF\), corresponding angles are equal: \(\angle P = \angle E\) and \(\angle Q = \angle F\).
5. Chain the equalities. From steps 2 and 4, \(\angle B = \angle P = \angle E\) and \(\angle C = \angle Q = \angle F\). Together with the given \(\angle A = \angle D\), all three pairs of corresponding angles match.
6. Conclude similarity. Why? — All three pairs of corresponding angles are equal, so the AAA similarity criterion applies and \(\Delta ABC \sim \Delta DEF\).

One detail the book leaves silent: fixing the ratio below 1 makes ABC the smaller triangle, so the cut points P and Q actually lie on the sides of DEF. If the ratio were greater than 1, swap the labels on the two triangles and run the same argument.

Where the book uses Theorem 6.5 after the proof

Two examples in the chapter apply the SAS criterion directly, and they show the two exam patterns to recognise. The table also names the contrast examples, so you can see why each criterion is chosen.

Example 6 (NCERT, p. 92) is the direct application. The given product equality \(OA \cdot OB = OC \cdot OD\) is rewritten as the ratio \(\frac{OA}{OC} = \frac{OD}{OB}\), the vertically opposite angles \(\angle AOD = \angle COB\) supply the equal included angle, and the SAS similarity criterion gives \(\Delta AOD \sim \Delta COB\).

Example 8 (NCERT, pp. 93–94) is the second pattern. Starting from \(\Delta ABC \sim \Delta PQR\), with CM and RN the medians, part (i) proves \(\Delta AMC \sim \Delta PNR\) by SAS similarity: the medians halve AB and PQ, so the side ratio survives, and the angle between the halved side and the adjacent side is already equal from the original similarity.

A quick worked check on fresh triangles

Take \(\Delta XYZ\) and \(\Delta UVW\) with \(XY = 4\text{ cm}\), \(XZ = 6\text{ cm}\), \(UV = 6\text{ cm}\), \(UW = 9\text{ cm}\), and \(\angle X = \angle U = 50^\circ\). Then \(\frac{XY}{UV} = \frac{4}{6} = \frac{2}{3}\) and \(\frac{XZ}{UW} = \frac{6}{9} = \frac{2}{3}\).

The equal angle sits between the two sides used in each triangle, so by the SAS similarity criterion \(\Delta XYZ \sim \Delta UVW\) with scale factor \(\frac{2}{3}\), which also gives \(\angle Y = \angle V\) and \(\angle Z = \angle W\).

Exercise 6.3 begins on page 94. Its first question shows pairs of triangles and asks you to state which are similar, name the similarity criterion used, and write the pairs in symbolic form — recognising SAS similarity is part of that task (NCERT, p. 94).

Related criteria: AAA, SSS and the theorems the proof leans on

Theorem 6.5 sits inside a chain of results from the same chapter. The Basic Proportionality Theorem (Thales Theorem) and its converse give the tool that justifies \(PQ \parallel EF\) inside this proof. The AAA similarity criterion is what the page 90 activity uses and what closes the proof. The SSS similarity criterion is the sibling that covers the sides-only route.

The chapter prints no separate converse for Theorem 6.5. The converse direction is immediate from what similar triangles mean — all corresponding angles equal and all sides proportional — a fact the book itself relies on in Example 8.

Choosing between AAA, SSS and SAS similarity

Mistakes that cost marks in SAS similarity questions

Questions students ask about the SAS similarity theorem

What is the SAS similarity theorem?

It is Theorem 6.5 of NCERT Class 10 Maths chapter 6 (Triangles): when one angle of a triangle equals one angle of another, and the sides including these angles are proportional, the two triangles are similar. The name SAS (Side-Angle-Side) describes the data used — two sides and the angle between them.

What is the difference between SAS similarity and SAS congruence?

Both need two sides and the included angle. Congruence insists the two sides are exactly equal, so the ratio is 1. Similarity accepts any equal ratio — a triangle enlarged to double size is similar, not congruent. Congruence is similarity with scale factor 1.

Why does the equal angle have to be included between the two proportional sides?

Because the proof places a copy of ABC inside DEF at the equal angle D, and the two sides that meet at D are the ones that carry the ratio. If the equal angle were not included, you would be comparing two sides and a non-included angle — the SSA case — which does not guarantee similarity.

How do you show that PQ is parallel to EF in the proof of Theorem 6.5?

By construction \(DP = AB\) and \(DQ = AC\), so the given ratio \(\frac{AB}{DE} = \frac{AC}{DF}\) becomes \(\frac{DP}{DE} = \frac{DQ}{DF}\). P and Q therefore divide DE and DF in the same ratio, and the converse of the basic proportionality theorem says the line joining them is parallel to the third side, EF.

Which questions in Exercise 6.3 ask you to use the SAS similarity criterion?

Question 1 of Exercise 6.3 (page 94) is the recogniser question: it shows pairs of triangles and asks you to state which are similar, name the criterion used, and write the pairs in symbolic form. SAS similarity is one of the criteria you must recognise there; which pairs use it depends on the measurements in the figure.

Match each pair against the Example 6 pattern — proportional sides plus an equal included angle.

Related resources

This listing is maintained for the 2026-27 academic session using the NCERT textbook information available to us. NCERT remains the authority for confirming the latest edition.

For the full chapter, open the Triangles 6 chapter page. For the wider set of Class 10 material, use the Class 10 Mathematics notes hub, the Class 10 section or the CBSE notes home. The sibling theorems of this chapter are the Basic Proportionality Theorem (Thales Theorem), its converse, the AAA similarity criterion and the SSS similarity criterion.

Other theorem pages on this site include proving root 2 is irrational and the tangent perpendicular to radius theorem.

The statement, figure and proof above follow the official edition of the NCERT Class 10 Mathematics textbook on ncert.nic.in, where Theorem 6.5 appears on printed page 90 of chapter 6 (Triangles) and the proof continues to page 91.

Textbook contents and the examinable syllabus are not always identical — check the current official CBSE syllabus for what is examined this session.

Reference: NCERT Class 10 Mathematics textbook, chapter 6 (Triangles), Theorem 6.5, official edition on ncert.nic.in.