Is it correct to say: Writing the similarity in the wrong vertex order?

Match equal angles first. If ∠A = ∠R and ∠B = ∠Q, write ΔABC ~ ΔRQP, and then AB/RQ = BC/QP = CA/PR.

Is it correct to say: Mixing part and whole in BPT, e.g. AD/AB = AE/EC?

Use part/part (AD/DB = AE/EC) or part/whole on both sides consistently (AD/AB = AE/AC). Never mix the two.

Is it correct to say: Using SAS with an angle that is not between the two sides?

The equal angle must be the INCLUDED angle: for AB/PQ = AC/PR you need ∠A = ∠P.

Is it correct to say: Calling two quadrilaterals similar just because their angles are equal?

For polygons other than triangles, both equal angles AND proportional sides are required. A 2 × 3 rectangle is not similar to a square.

Is it correct to say: Thinking similar triangles must be congruent?

Similar means same shape; size can differ. Congruent triangles are the special case where the ratio of sides is 1.

Is it correct to say: Treating the shadow tip as the person's position in lamp-post problems?

The big triangle's base is (distance from the post + shadow length), not just the distance from the post.

📄 Formula sheet (PDF)

Triangles

Similar figures, similarity of triangles, Basic Proportionality Theorem and its converse, and the AAA, AA, SSS and SAS similarity criteria — NCERT Class 10 Maths Ch 6

📐 14 formulas✏️ 4 examples🎯 6 practice⚖️ Unit IV (Geometry): 15 marks (with Ch 10)🏫 CBSE📚 Class 10✓ 2025–26 syllabus

💡

Board Exam Tips

→The Basic Proportionality Theorem (BPT) is the theorem the syllabus lists for proof — practise its area-based proof with a neat labelled figure.
→Write similarity with matching vertex order: if ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R, write ΔABC ~ ΔPQR, not ΔABC ~ ΔQPR. The ratios are read from this order.
→In every similarity proof, give a reason for each equal angle (common angle, vertically opposite angles, alternate angles) and name the criterion used (AA, SSS or SAS).
→For SAS, the equal angle must be the one INCLUDED between the two proportional sides.
→Shadow and lamp-post problems are AA similarity in disguise: draw the two right triangles and mark the equal angle made by the light ray.

📐 Formulas(14)

Similar Polygons

Similar ⟺ corresponding angles equal and \frac{A B}{P Q} = \frac{B C}{QR} = \dots

Similar Triangles (Definition)

△ A B C \sim △ P QR ⟺ ∠ A = ∠ P, ∠ B = ∠ Q, ∠ C = ∠ R and \frac{A B}{P Q} = \frac{B C}{QR} = \frac{C A}{R P}

Basic Proportionality Theorem (Thales)★ Board fav

D E ∥ B C \Rightarrow \frac{A D}{D B} = \frac{A E}{E C}

Symbol	Meaning
$D$	Point where the parallel line cuts side AB
$E$	Point where the parallel line cuts side AC

BPT — Other Useful Forms

D E ∥ B C \Rightarrow \frac{A D}{A B} = \frac{A E}{A C}, \frac{D B}{A B} = \frac{E C}{A C}

Converse of BPT★ Board fav

\frac{A D}{D B} = \frac{A E}{E C} \Rightarrow D E ∥ B C

AAA Similarity Criterion

∠ A = ∠ P, ∠ B = ∠ Q, ∠ C = ∠ R \Rightarrow △ A B C \sim △ P QR

AA Similarity Criterion★ Board fav

∠ A = ∠ P, ∠ B = ∠ Q \Rightarrow △ A B C \sim △ P QR

SSS Similarity Criterion★ Board fav

\frac{A B}{P Q} = \frac{B C}{QR} = \frac{C A}{R P} \Rightarrow △ A B C \sim △ P QR

SAS Similarity Criterion★ Board fav

∠ A = ∠ P, \frac{A B}{P Q} = \frac{A C}{P R} \Rightarrow △ A B C \sim △ P QR

Ratio of Perimeters

\frac{Perimeter of △ A B C}{Perimeter of △ P QR} = \frac{A B}{P Q} = \frac{B C}{QR} = \frac{C A}{R P}

Corresponding Medians and Altitudes

△ A B C \sim △ P QR \Rightarrow \frac{A D}{P M} = \frac{A B}{P Q}

Diagonals of a Trapezium

A B ∥ D C \Rightarrow \frac{O A}{O C} = \frac{O B}{O D} = \frac{A B}{C D}

Shadows at the Same Time

\frac{h _{1}}{s _{1}} = \frac{h _{2}}{s _{2}}

Symbol	Meaning
$h_{1}, h_{2}$	Heights of the two vertical objects (m)
$s_{1}, s_{2}$	Lengths of their shadows (m)

Lamp-post and Shadow

\frac{H}{d + s} = \frac{h}{s}

Symbol	Meaning
$H$	Height of the lamp above the ground (m)
$h$	Height of the person (m)
$d$	Distance of the person from the foot of the lamp-post (m)
$s$	Length of the person's shadow (m)

✏️ Solved Examples

1Solved Exampleeasy3 steps

In ΔABC, D and E are points on AB and AC such that DE ∥ BC. If AD = 3 cm, DB = 5 cm and AE = 4.5 cm, find EC.

Since DE ∥ BC, apply the Basic Proportionality Theorem.

\frac{A D}{D B} = \frac{A E}{E C}

2Solved Exampleboard3 steps

In ΔABC, D lies on AB and E lies on AC with AD = 4 cm, DB = 6 cm, AE = 5 cm and EC = 7.5 cm. Is DE ∥ BC?

Find the ratio in which D divides AB.

\frac{A D}{D B} = \frac{4}{6} = \frac{2}{3}

3Solved Exampleboard4 steps

ABCD is a trapezium with AB ∥ DC. Its diagonals AC and BD intersect at O. Prove that OA/OC = OB/OD.

Consider ΔAOB and ΔCOD. With AB ∥ DC and AC as transversal, alternate angles are equal.

∠ O A B = ∠ O C D

4Solved ExampleHOTS4 steps

A boy 1.6 m tall walks away from the foot of a lamp-post at 1.5 m/s. The lamp is 4.8 m above the ground. Find the length of his shadow after 4 seconds.

Distance walked in 4 s.

d = 1.5 \times 4 = 6 m

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Writing the similarity in the wrong vertex order
✓Match equal angles first. If ∠A = ∠R and ∠B = ∠Q, write ΔABC ~ ΔRQP, and then AB/RQ = BC/QP = CA/PR.
2
Mixing part and whole in BPT, e.g. AD/AB = AE/EC
✓Use part/part (AD/DB = AE/EC) or part/whole on both sides consistently (AD/AB = AE/AC). Never mix the two.
3
Using SAS with an angle that is not between the two sides
✓The equal angle must be the INCLUDED angle: for AB/PQ = AC/PR you need ∠A = ∠P.
4
Calling two quadrilaterals similar just because their angles are equal
✓For polygons other than triangles, both equal angles AND proportional sides are required. A 2 × 3 rectangle is not similar to a square.
5
Thinking similar triangles must be congruent
✓Similar means same shape; size can differ. Congruent triangles are the special case where the ratio of sides is 1.
6
Treating the shadow tip as the person's position in lamp-post problems
✓The big triangle's base is (distance from the post + shadow length), not just the distance from the post.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
In ΔABC, DE ∥ BC with D on AB and E on AC. If AD = 2.4 cm, AB = 6 cm and AC = 8 cm, find AE.
Q2
In ΔPQR, S and T lie on PQ and PR with PS = 3 cm, SQ = 5 cm, PT = 4 cm and TR = 6 cm. Is ST ∥ QR?
Q3
ABCD is a trapezium with AB ∥ DC and diagonals meeting at O. If OA = 6 cm, OC = 4 cm and OB = 9 cm, find OD.
Q4
A vertical pole 7 m high casts a shadow 5 m long. At the same time a building casts a shadow 35 m long. Find the height of the building.
Q5
ΔABC ~ ΔPQR. The perimeters of ΔABC and ΔPQR are 36 cm and 24 cm. If PQ = 10 cm, find AB.
Q6
In ΔABC and ΔDEF, ∠B = ∠E = 50°, AB = 4 cm, BC = 6 cm, DE = 6 cm and EF = 9 cm. Are the triangles similar? Name the criterion.

📝 Notes

Triangles

Two figures are similar if they have the same shape but not necessarily the same size. All circles are similar, all squares are similar and all equilateral triangles are similar. Congruence is the special case where the scale factor is 1.

Why triangles are special

For general polygons you must check BOTH equal angles and proportional sides. For triangles, either condition gives the other, which is why the criteria are short:

AA: two pairs of equal angles.
SSS: all three side ratios equal.
SAS: one equal angle with the sides including it in the same ratio.

Once you know ΔABC ~ ΔPQR, every corresponding length is in the same ratio: sides, perimeters, medians and altitudes.

BPT: the workhorse

The Basic Proportionality Theorem says that a line parallel to one side cuts the other two sides in the same ratio: if DE ∥ BC, then AD/DB = AE/EC. Its converse is how you prove two lines parallel. When D is the mid-point of AB, BPT shows that E is the mid-point of AC, which is the converse of the Class 9 mid-point theorem.

How to attack a proof

Mark the given parallel lines and look for alternate or corresponding angles.
Look for a common angle or vertically opposite angles shared by two triangles.
Write the two triangles with matching vertex order.
Name the criterion and then write the ratio you need.

Real-life similarity

Shadow problems use the fact that, at the same moment, the sun's rays make the same angle with level ground. A pole and a tower (or a person and a lamp-post) form two right triangles with an equal angle, so they are similar by AA. Set up height/shadow on both sides and solve.

Note: in the rationalised NCERT book (2023 onwards), the theorem on areas of similar triangles and the Pythagoras theorem section have been removed from this chapter.