Is it correct to say: Writing BPT as AD/AB = AE/EC (a part over the whole on one side, a part over a part on the other)?

Keep both ratios of the same type: AD/DB = AE/EC (part : part) or AD/AB = AE/AC (part : whole).

Is it correct to say: Inverting the angle-bisector ratio as BD/DC = AC/AB?

Each part of the base goes with the side next to it: BD is adjacent to AB, so BD/DC = AB/AC.

Is it correct to say: Taking the ratio of areas of similar triangles equal to the ratio of their sides?

Areas go as the SQUARE of the ratio of corresponding sides. Sides 3 : 5 ⇒ areas 9 : 25.

Is it correct to say: Using 'ratio of areas = ratio of bases' when the heights are not equal?

That shortcut needs equal heights (e.g. a common vertex with bases on one line). Otherwise use the product of base and height.

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

The equal angle must be the included angle. For example, AB/PQ = AC/PR needs ∠A = ∠P.

Is it correct to say: Writing ΔABC ~ ΔQPR when A matches P and B matches Q?

The letter order shows the correspondence. Write the matching vertices in the same positions: ΔABC ~ ΔPQR.

📄 Formula sheet (PDF)

Similarity

Ratio of areas of triangles, basic proportionality theorem, property of an angle bisector, three parallel lines, similarity tests and areas of similar triangles — Maharashtra SSC Geometry Ch 1

📐 14 formulas✏️ 4 examples🎯 6 practice🏫 Maharashtra State Board📚 Class 10✓ 2025–26 syllabus

💡

Board Exam Tips

→The proofs of the Basic Proportionality Theorem and the theorem of areas of similar triangles are standard textbook theorems. Practise writing them with figure, Given, To prove, Construction and Proof.
→Write the correspondence of vertices first (ΔABC ~ ΔPQR) and then the ratios in the same order. Pairing the wrong sides is a common source of wrong answers.
→When a question asks for a ratio of areas, first check what is common: equal heights → ratio of bases; equal bases → ratio of heights; similar triangles → ratio of squares of sides.
→In BPT and angle-bisector sums, write the proportion symbolically (AD/DB = AE/EC) before substituting numbers. It earns a step mark and prevents mixing a part with the whole.
→Name the similarity test (AA, SAS or SSS) in your answer — 'by AA test of similarity' is a required reason, not decoration.

📐 Formulas(14)

Ratio of Areas of Two Triangles★ Board fav

\frac{A ( △ A B C )}{A ( △ P QR )} = \frac{B C \times A D}{QR \times P S}

Symbol	Meaning
$A (△ A B C)$	Area of triangle ABC (cm²)
$B C, QR$	Bases of the two triangles (cm)
$A D, P S$	Heights drawn to those bases (cm)

Triangles with Equal Heights

\frac{A _{1}}{A _{2}} = \frac{b _{1}}{b _{2}}

Symbol	Meaning
$A_{1}, A_{2}$	Areas of the two triangles
$b_{1}, b_{2}$	Their corresponding bases

Triangles with Equal Bases

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

Symbol	Meaning
$A_{1}, A_{2}$	Areas of the two triangles
$h_{1}, h_{2}$	Their corresponding heights

Basic Proportionality Theorem (BPT)★ Board fav

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

Symbol	Meaning
$D, E$	Points on sides AB and AC where the parallel line cuts them

Other Forms of BPT

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

Converse of BPT

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

Property of an Angle Bisector of a Triangle★ Board fav

∠ B A D = ∠ D A C \Rightarrow \frac{A B}{A C} = \frac{B D}{D C}

Symbol	Meaning
$A D$	Bisector of ∠BAC meeting BC at D
$B D, D C$	Parts of side BC

Converse of the Angle Bisector Property

\frac{A B}{A C} = \frac{B D}{D C} \Rightarrow ray A D bisects ∠ B A C

Property of Three Parallel Lines and their Transversals

l ∥ m ∥ n \Rightarrow \frac{A B}{B C} = \frac{P Q}{QR}

Symbol	Meaning
$A B, B C$	Intercepts on the first transversal
$P Q, QR$	Corresponding intercepts on the second transversal

Similar Triangles

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

AAA / AA Test of Similarity

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

SAS Test of Similarity

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

SSS Test of Similarity

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

Theorem of Areas of Similar Triangles★ Board fav

△ A B C \sim △ P QR \Rightarrow \frac{A ( △ A B C )}{A ( △ P QR )} = \frac{A B ^{2}}{P Q ^{2}} = \frac{B C ^{2}}{Q R ^{2}} = \frac{A C ^{2}}{P R ^{2}}

Symbol	Meaning
$A B, P Q$	A pair of corresponding sides

✏️ Solved Examples

1Solved Exampleeasy2 steps

The base and height of ΔABC are 9 cm and 5 cm. The base and height of ΔPQR are 10 cm and 6 cm. Find A(ΔABC) : A(ΔPQR).

Ratio of areas = ratio of products of base and height.

\frac{A ( △ A B C )}{A ( △ P QR )} = \frac{9 \times 5}{10 \times 6}

2Solved Exampleboard3 steps

In ΔABC, the bisector of ∠A meets side BC at D. If AB = 6 cm, AC = 9 cm and BC = 10 cm, find BD and DC.

By the property of an angle bisector of a triangle:

\frac{B D}{D C} = \frac{A B}{A C} = \frac{6}{9} = \frac{2}{3}

3Solved Exampleboard3 steps

ΔABC ~ ΔPQR. A(ΔABC) = 80 cm², A(ΔPQR) = 125 cm² and AB = 12 cm. Find PQ.

By the theorem of areas of similar triangles:

\frac{A ( △ A B C )}{A ( △ P QR )} = \frac{A B ^{2}}{P Q ^{2}}

4Solved ExampleHOTS4 steps

In trapezium ABCD, side AB ∥ side DC, and the diagonals AC and BD intersect at O. If AB = 18 cm, DC = 12 cm and OD = 8 cm, find OB and A(ΔAOB) : A(ΔCOD).

∠AOB = ∠COD (vertically opposite angles) and ∠OAB = ∠OCD (alternate angles, AB ∥ DC with transversal AC). So the triangles are similar by AA.

△ A O B \sim △ C O D

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Writing BPT as AD/AB = AE/EC (a part over the whole on one side, a part over a part on the other)
✓Keep both ratios of the same type: AD/DB = AE/EC (part : part) or AD/AB = AE/AC (part : whole).
2
Inverting the angle-bisector ratio as BD/DC = AC/AB
✓Each part of the base goes with the side next to it: BD is adjacent to AB, so BD/DC = AB/AC.
3
Taking the ratio of areas of similar triangles equal to the ratio of their sides
✓Areas go as the SQUARE of the ratio of corresponding sides. Sides 3 : 5 ⇒ areas 9 : 25.
4
Using 'ratio of areas = ratio of bases' when the heights are not equal
✓That shortcut needs equal heights (e.g. a common vertex with bases on one line). Otherwise use the product of base and height.
5
Applying SAS similarity with an angle that is not between the two proportional sides
✓The equal angle must be the included angle. For example, AB/PQ = AC/PR needs ∠A = ∠P.
6
Writing ΔABC ~ ΔQPR when A matches P and B matches Q
✓The letter order shows the correspondence. Write the matching vertices in the same positions: ΔABC ~ ΔPQR.

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Two triangles have equal heights and bases 12 cm and 8 cm. Find the ratio of their areas.
Q2
In ΔABC, D is on AB and E is on AC with DE ∥ BC. If AD = 5 cm, DB = 7.5 cm and AE = 4 cm, find EC.
Q3
In ΔPQR, ray PS bisects ∠QPR and meets QR at S. If PQ = 7 cm, PR = 5 cm and QS = 3.5 cm, find SR.
Q4
Lines l ∥ m ∥ n cut transversal t₁ at A, B, C and transversal t₂ at P, Q, R. If AB = 3 cm, BC = 4.5 cm and PR = 10 cm, find PQ and QR.
Q5
ΔLMN ~ ΔXYZ. A(ΔLMN) = 64 cm², A(ΔXYZ) = 100 cm² and LM = 12 cm. Find XY.
Q6
The sides of one triangle are 4 cm, 6 cm, 8 cm and of another are 6 cm, 9 cm, 12 cm. Are the triangles similar? If so, by which test, and what is the ratio of their areas?

📝 Notes

Similarity

Similarity is the first chapter of Geometry, and the rest of the book leans on it: the proof of Pythagoras theorem in Chapter 2 and many circle results in Chapter 3 are similarity arguments.

Areas are the engine of the proofs

The textbook proves the Basic Proportionality Theorem using areas, not lengths, and suggests an area proof of the angle-bisector property as an alternative to its BPT-based proof. Remember the three area facts:

Ratio of areas = ratio of (base × height).
Equal heights ⇒ ratio of areas = ratio of bases.
Equal bases ⇒ ratio of areas = ratio of heights.

Once these are fluent, the proof of BPT is three lines: two "equal height" ratios, plus the fact that triangles on the same base between the same parallels have equal areas.

Which result to use

Given in the question	Use
A line parallel to a side	BPT: $\frac{AD}{DB} = \frac{AE}{EC}$
A ratio of parts, asked to prove lines parallel	Converse of BPT
An angle bisector meeting the opposite side	$\frac{AB}{AC} = \frac{BD}{DC}$
Three parallel lines and two transversals	Ratio of intercepts is the same
Two triangles with equal angles or proportional sides	AA, SAS or SSS test
Areas of similar triangles	Ratio of squares of corresponding sides

Writing a similarity proof

Name the two triangles and list the equal angles or proportional sides with reasons (vertically opposite angles, alternate angles, common angle, given).
Name the test: AA, SAS or SSS.
Write the correspondence with vertices in matching order, for example $\triangle AOB \sim \triangle COD$ .
Only then write the proportion of corresponding sides and solve.

Quick checks

If the area ratio you get is the same as the side ratio, you have forgotten to square.
In an angle-bisector sum, the longer adjacent side always gets the longer part of the base.
Similar triangles need not be congruent — congruent triangles are similar with ratio 1.