Topic 1: Congruence vs Similarity of Geometric Figures
Core Axioms
Congruence ($\cong$): Same shape and same size. Corresponding sides are strictly equal ($AB = DE, BC = EF, AC = DF$).
Similarity ($\sim$): Same shape, but not necessarily the same size.
• All corresponding angles are equal ($\angle A = \angle D, \angle B = \angle E, \angle C = \angle F$).
• All corresponding sides are proportional ($\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = k$).
Universal Golden Rule:All congruent figures are similar, but similar figures need NOT be congruent.
Always Similar Figures: All circles, all squares, all equilateral triangles, and all regular polygons of $n$ sides.
Topic 1 Practice Kit: Similarity Concepts & Polygon Conditions
Problem 1.1 (Fill in the Blanks / 1 Mark) NCERT EX 6.1
(i) All circles are ________. (congruent / similar)
(ii) All squares are ________. (similar / congruent)
(iii) All ________ triangles are similar. (isosceles / equilateral)
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are ________ and (b) their corresponding sides are ________.
Answers & Key Rationale:
(i) similar (same circular shape, radii may vary).
(ii) similar (each angle is $90^\circ$, sides are proportional).
(iii) equilateral (all internal angles are strictly $60^\circ$).
(iv) (a) equal, (b) proportional.
Problem 1.2 (Square vs Rhombus Similarity Test) NCERT EX 6.1 Q3 / RD SHARMA
State whether a square of side $3\text{ cm}$ and a rhombus of side $3\text{ cm}$ (with acute angle $60^\circ$) are similar. Give a valid mathematical reason.
Step-by-Step Solution:
Ratio of corresponding sides $= \frac{3}{3} = 1$ (sides are in proportion).
However, in the square, all angles are $90^\circ$, whereas in the rhombus, angles are $60^\circ$ and $120^\circ$.
Since their corresponding angles are NOT equal, the square and rhombus are NOT similar.
Statement: If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
Fig 6.2: Basic Proportionality Theorem (BPT) with Altitudes $DM \perp AC$ and $EN \perp AB$
Given: In $\Delta ABC$, line $DE \parallel BC$ intersects $AB$ at $D$ and $AC$ at $E$.
To Prove: $\frac{AD}{DB} = \frac{AE}{EC}$.
Construction: Join $B$ to $E$ and $C$ to $D$. Draw $EN \perp AB$ and $DM \perp AC$.
Formal Proof:
$\text{Area}(\Delta ADE) = \frac{1}{2} \times AD \times EN$ and $\text{Area}(\Delta BDE) = \frac{1}{2} \times DB \times EN$.
$$\implies \frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta BDE)} = \frac{AD}{DB} \quad \text{--- (1)}$$
$\Delta BDE$ and $\Delta DEC$ lie on the same base $DE$ and between the same parallel lines $DE \parallel BC$:
$$\implies \text{Area}(\Delta BDE) = \text{Area}(\Delta DEC) \quad \text{--- (3)}$$
From equations (1), (2), and (3), we get:
$$\mathbf{\frac{AD}{DB} = \frac{AE}{EC}} \qquad \text{(Hence Proved!)} \quad \blacksquare$$
Problem 2.3 (Nested Parallel Lines Chain: AD² = AB × AF) CBSE 2019 / RD SHARMA
In $\Delta ABC$, a line $DE \parallel BC$ intersects $AB$ at $D$ and $AC$ at $E$. A line $EF \parallel CD$ intersects $AB$ at $F$. Prove that $\mathbf{AD^2 = AB \times AF}$.
Step-by-Step Solution:
In $\Delta ABC$, $DE \parallel BC \implies \frac{AD}{AB} = \frac{AE}{AC}$ … (I).
In $\Delta ADC$, $FE \parallel DC \implies \frac{AF}{AD} = \frac{AE}{AC}$ … (II).
From (I) and (II): $\frac{AD}{AB} = \frac{AF}{AD} \implies \mathbf{AD^2 = AB \times AF}$. ■
$DE \parallel BC$ and $FE \parallel DC$
Topic 3: Converse of Basic Proportionality Theorem (Theorem 6.2)
Theorem 6.2: Converse of BPT
Statement: If a line divides any two sides of a triangle in the same ratio, then the line is parallel to the third side.
Fig 6.3: Converse of BPT — Proving $E'$ Coincides with $E$
Proof (By Contradiction):
Suppose $DE$ is NOT parallel to $BC$. Then draw a line $DE' \parallel BC$ meeting $AC$ at $E'$.
By BPT on $\Delta ABC$ with $DE' \parallel BC$: $\frac{AD}{DB} = \frac{AE'}{E'C}$.
Adding $1$ to both sides: $\frac{AE' + E'C}{E'C} = \frac{AE + EC}{EC} \implies \frac{AC}{E'C} = \frac{AC}{EC} \implies \mathbf{E'C = EC}$.
This is only possible if points $E'$ and $E$ coincide. Therefore, $\mathbf{DE \parallel BC}$. ■
Topic 3 Practice Kit: Testing Parallelism (Converse of BPT)
Problem 3.1 (Checking Parallelism in Triangles) NCERT EX 6.2 Q2
$E$ and $F$ are points on sides $PQ$ and $PR$ of $\Delta PQR$. State whether $EF \parallel QR$ in each case:
(a) $PE = 3.9\text{ cm}, EQ = 3\text{ cm}, PF = 3.6\text{ cm}, FR = 2.4\text{ cm}$.
(b) $PE = 4\text{ cm}, QE = 4.5\text{ cm}, PF = 8\text{ cm}, RF = 9\text{ cm}$.
Step-by-Step Solution:
Case (a): $\frac{PE}{EQ} = \frac{3.9}{3} = 1.3$ and $\frac{PF}{FR} = \frac{3.6}{2.4} = 1.5$.
Since $\frac{PE}{EQ} \ne \frac{PF}{FR} \implies \mathbf{EF \text{ is NOT parallel to } QR}$.
Case (b): $\frac{PE}{QE} = \frac{4}{4.5} = \frac{8}{9}$ and $\frac{PF}{RF} = \frac{8}{9}$.
Since $\frac{PE}{QE} = \frac{PF}{RF} = \frac{8}{9} \implies \mathbf{EF \parallel QR}$ (By Converse of BPT).
Problem 3.2 (Converse of BPT in Quadrilaterals) RD SHARMA
In $\Delta ABC$, $D$ and $E$ are points on $AB$ and $AC$ such that $AD = x, DB = x - 2, AE = x + 2$, and $EC = x - 1$. Find the value of $x$ for which $DE \parallel BC$.
Trapezium Diagonals Property: The diagonals of a trapezium with $AB \parallel DC$ intersect at $O$ such that:
$$\mathbf{\frac{AO}{BO} = \frac{CO}{DO} \iff \frac{AO}{CO} = \frac{BO}{DO}}$$
Converse Property: If the diagonals of a quadrilateral divide each other proportionally, then it is a trapezium.
Line Parallel to Bases: If $EF \parallel AB \parallel DC$ in trapezium $ABCD$ intersecting non-parallel sides at $E$ and $F$, then $\mathbf{\frac{AE}{ED} = \frac{BF}{FC}}$.
Theorem: Trapezium Diagonals Proportionality
Statement: In trapezium $ABCD$ with $AB \parallel DC$, diagonals $AC$ and $BD$ intersect at $O$. Prove that $\frac{AO}{BO} = \frac{CO}{DO}$.
Fig 6.4: Trapezium $ABCD$ with Diagonals Intersecting at $O$ and Construction $OE \parallel AB \parallel DC$
Proof: Draw $OE \parallel AB \parallel DC$ meeting $AD$ at $E$.
In $\Delta ADC$, $OE \parallel DC \implies \frac{AE}{ED} = \frac{AO}{OC}$ (by BPT).
In $\Delta DAB$, $OE \parallel AB \implies \frac{DE}{EA} = \frac{DO}{OB} \implies \frac{AE}{ED} = \frac{BO}{OD}$.
Problem 4.2 (Proving Quadrilateral is a Trapezium) NCERT EX 6.2 Q10
The diagonals of a quadrilateral $ABCD$ intersect each other at the point $O$ such that $\frac{AO}{BO} = \frac{CO}{DO}$. Show that $ABCD$ is a trapezium.
Step-by-Step Solution:
Given: $\frac{AO}{CO} = \frac{BO}{DO}$. Draw $OE \parallel AB$ meeting $AD$ at $E$.
In $\Delta DAB$, $OE \parallel AB \implies \frac{DE}{EA} = \frac{DO}{OB} \implies \frac{AE}{ED} = \frac{BO}{DO} = \frac{AO}{CO}$.
In $\Delta ADC$, $\frac{AE}{ED} = \frac{AO}{CO} \implies OE \parallel DC$ (by Converse of BPT).
Since $OE \parallel AB$ and $OE \parallel DC \implies \mathbf{AB \parallel DC}$.
Therefore, $ABCD$ is a trapezium. ■
Problem 4.3 (Trapezium Central Segment: OE = OF) 5 MARKS / RS AGGARWAL
Diagonals of a trapezium $ABCD$ with $AB \parallel DC$ intersect at $O$. A line through $O$ parallel to $AB$ intersects $AD$ at $E$ and $BC$ at $F$. Prove that $OE = OF$.
Step-by-Step Solution:
In $\Delta ADC$, $OE \parallel DC \implies \frac{OE}{DC} = \frac{AO}{AC}$.
In $\Delta BDC$, $OF \parallel DC \implies \frac{OF}{DC} = \frac{BO}{BD}$.
Since $\frac{AO}{AC} = \frac{BO}{BD}$ (by trapezium diagonals proportionality), we get $\frac{OE}{DC} = \frac{OF}{DC} \implies \mathbf{OE = OF}$. ■
Topic 5: Criteria for Similarity of Triangles (AA, SSS, SAS)
Theorem 6.3: AA (Angle-Angle) Similarity Criterion
Statement: If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.
$$\text{If } \angle A = \angle D \text{ and } \angle B = \angle E \implies \mathbf{\Delta ABC \sim \Delta DEF}$$
Fig 6.5: AA Similarity — Two Equal Corresponding Angles Guarantee Equiangular Triangles
SSS & SAS Similarity Criteria
SSS Similarity (Theorem 6.4): If all 3 corresponding sides are proportional:
$$\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} \implies \mathbf{\Delta ABC \sim \Delta DEF}$$
SAS Similarity (Theorem 6.5): If 2 pairs of sides are proportional and the included angle is equal:
$$\frac{AB}{DE} = \frac{AC}{DF} \quad \text{and} \quad \angle A = \angle D \implies \mathbf{\Delta ABC \sim \Delta DEF}$$
Mandatory Teacher's Rule on SAS
The equal angle MUST strictly be between the two proportional sides. If $\frac{AB}{DE} = \frac{AC}{DF}$, but $\angle B = \angle E$ is given, SAS similarity CANNOT be applied!
Topic 5 Practice Kit: Criteria for Similarity (NCERT / Board PYQs)
Problem 5.1 (NCERT Classic Ratio & Angle) NCERT EX 6.3 Q4
In the figure, $\frac{QR}{QS} = \frac{QT}{PR}$ and $\angle 1 = \angle 2$. Show that $\Delta PQS \sim \Delta TQR$.
Step-by-Step Solution:
In $\Delta PQR$, $\angle 1 = \angle 2 \implies PQ = PR$.
Substitute $PR = PQ$ in given ratio: $\frac{QR}{QS} = \frac{QT}{PQ} \implies \frac{PQ}{QT} = \frac{QS}{QR}$.
In $\Delta PQS$ and $\Delta TQR$, $\frac{PQ}{QT} = \frac{QS}{QR}$ and $\angle Q = \angle Q$ (common).
Therefore, by SAS Similarity, $\mathbf{\Delta PQS \sim \Delta TQR}$. ■
Problem 5.2 (Side-Square Similarity Proof) NCERT EX 6.3 Q13 / CBSE 2023
$D$ is a point on side $BC$ of $\Delta ABC$ such that $\angle ADC = \angle BAC$. Prove that $CA^2 = CB \times CD$.
Step-by-Step Solution:
In $\Delta ABC$ and $\Delta DAC$: $\angle BAC = \angle ADC$ (given) and $\angle C = \angle C$ (common).
By AA similarity, $\Delta ABC \sim \Delta DAC$.
$$\frac{CA}{CD} = \frac{CB}{CA} \implies \mathbf{CA^2 = CB \times CD} \quad \blacksquare$$
Problem 5.3 (Isosceles Triangle Perpendiculars) NCERT EX 6.3 Q11
In figure, $E$ is a point on side $CB$ produced of an isosceles triangle $ABC$ with $AB = AC$. If $AD \perp BC$ and $EF \perp AC$, prove that $\Delta ABD \sim \Delta ECF$.
Step-by-Step Solution:
Since $AB = AC \implies \angle B = \angle C$.
In $\Delta ABD$ and $\Delta ECF$:
• $\angle ADB = \angle EFC = 90^\circ$ (Perpendiculars).
• $\angle ABD = \angle ECF$ (Angles opposite to equal sides).
Therefore, by AA Similarity Criterion, $\mathbf{\Delta ABD \sim \Delta ECF}$. ■
$\Delta ABD \sim \Delta ECF$ by AA
Problem 5.4 (Intersecting Altitudes in Acute Triangle) NCERT EX 6.3 Q7
Altitudes $AD$ and $CE$ of $\Delta ABC$ intersect each other at the point $P$. Show that:
(i) $\Delta AEP \sim \Delta CDP$, (ii) $\Delta ABD \sim \Delta CBE$.
Step-by-Step Solution:
(i) In $\Delta AEP$ and $\Delta CDP$: $\angle AEP = \angle CDP = 90^\circ$ and $\angle APE = \angle CPD$ (Vertically opposite angles) $\implies \mathbf{\Delta AEP \sim \Delta CDP}$ (AA).
(ii) In $\Delta ABD$ and $\Delta CBE$: $\angle ADB = \angle CEB = 90^\circ$ and $\angle B = \angle B$ (Common angle) $\implies \mathbf{\Delta ABD \sim \Delta CBE}$ (AA). ■
Example 6.1: The Sun Shadow Method NCERT EX 6.3 Q15
A vertical pole of length $6\text{ m}$ casts a shadow $4\text{ m}$ long on the ground and at the same time a tower casts a shadow $28\text{ m}$ long. Find the height of the tower.
Fig 6.6: Similar Right Triangles Formed by Sun Rays at the Same Instant
Solution: Since sun elevation angle is equal at the same instant, $\Delta ABC \sim \Delta PQR$ (AA Similarity).
$$\frac{\text{Height of Pole}}{\text{Height of Tower}} = \frac{\text{Shadow of Pole}}{\text{Shadow of Tower}} \implies \frac{6}{h} = \frac{4}{28} \implies h = 6 \times 7 = \mathbf{42\text{ metres}}$$
Topic 6 Practice Kit: Shadows, Lamp-Posts & Mirror Reflection
Problem 6.1 (Moving Person Lamp-Post Shadow) NCERT EXAMPLE 7 / CBSE 2023
A girl of height $90\text{ cm}$ walks away from the base of a lamp-post of height $3.6\text{ m}$ at a speed of $1.2\text{ m/s}$. Find the length of her shadow after 4 seconds.
Problem 6.2 (Ground Mirror Reflection Surveying) CBSE CASE STUDY / NCERT EXEMPLAR
A surveyor of height $1.8\text{ m}$ places a flat mirror on the ground $12\text{ m}$ from the base of a tall building. When he steps back $2.4\text{ m}$ from the mirror, he sees the top of the building reflected in the mirror. Calculate the height of the building.
Step-by-Step Solution:
By the Law of Reflection, $\text{Angle of Incidence} = \text{Angle of Reflection}$.
Since the person and the building stand vertically ($\angle = 90^\circ$), the triangles are similar by AA Similarity.
$$\frac{\text{Height of Building}}{\text{Height of Person}} = \frac{\text{Distance from Mirror to Building}}{\text{Distance from Mirror to Person}} \implies \frac{h}{1.8} = \frac{12}{2.4} = 5$$
$$\mathbf{h = 5 \times 1.8 = 9.0\text{ metres}}$$
Theorem 7.1: Proportionality of Medians & Sides (NCERT Ex 6.3 Q12)
Statement: Sides $AB, BC$ and median $AD$ of $\Delta ABC$ are proportional to sides $PQ, QR$ and median $PM$ of $\Delta PQR$. Prove that $\Delta ABC \sim \Delta PQR$.
Proof:
Given: $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$. Since $AD$ and $PM$ are medians, $BD = \frac{1}{2} BC$ and $QM = \frac{1}{2} QR$.
$\implies \angle B = \angle Q$ (corresponding angles of similar triangles).
In $\Delta ABC$ and $\Delta PQR$, $\frac{AB}{PQ} = \frac{BC}{QR}$ and $\angle B = \angle Q \implies \mathbf{\Delta ABC \sim \Delta PQR}$ (by SAS Similarity). ■
Theorem 7.2: The Hard Median Proportionality (NCERT Ex 6.3 Q14 — 5 MARKS)
Statement: Sides $AB, AC$ and median $AD$ of $\Delta ABC$ are proportional to sides $PQ, PR$ and median $PM$ of $\Delta PQR$. Prove that $\Delta ABC \sim \Delta PQR$.
Proof (Construction Method):
Produce $AD$ to $E$ such that $AD = DE$, and produce $PM$ to $N$ such that $PM = MN$. Join $EC$ and $NR$.
$\Delta ADB \cong \Delta EDC \implies AB = EC$. Similarly, $PQ = NR$.
Similarly, by joining $EB$ and $NQ$, we get $\angle DAB = \angle MPQ$.
Adding both equations: $\angle BAC = \angle QPR$.
Now in $\Delta ABC$ and $\Delta PQR$, $\frac{AB}{PQ} = \frac{AC}{PR}$ and $\angle A = \angle P \implies \mathbf{\Delta ABC \sim \Delta PQR}$ (by SAS Similarity). ■
Topic 7 Practice Kit: Medians, Altitudes & Perimeter Ratios
Problem 7.1 (Perimeter & Side Ratio in Similar Triangles) CBSE 2020 / RS AGGARWAL
$\Delta ABC \sim \Delta DEF$. If the perimeter of $\Delta ABC = 30\text{ cm}$, perimeter of $\Delta DEF = 18\text{ cm}$, and $BC = 9\text{ cm}$, find the length of $EF$.
Step-by-Step Solution:
Ratio of perimeters of similar triangles equals ratio of corresponding sides:
$$\frac{\text{Perimeter}(\Delta ABC)}{\text{Perimeter}(\Delta DEF)} = \frac{BC}{EF} \implies \frac{30}{18} = \frac{9}{EF}$$
$$EF = \frac{9 \times 18}{30} = \frac{162}{30} = \mathbf{5.4\text{ cm}}$$
Problem 7.2 (Altitudes Proportionality Proof) NCERT EX 6.3 Q16
If $AD$ and $PM$ are medians of triangles $ABC$ and $PQR$ respectively, where $\Delta ABC \sim \Delta PQR$, prove that: $$\frac{AB}{PQ} = \frac{AD}{PM}$$
Step-by-Step Solution:
Since $\Delta ABC \sim \Delta PQR \implies \frac{AB}{PQ} = \frac{BC}{QR}$ and $\angle B = \angle Q$.
Since $AD$ and $PM$ are medians, $BD = BC/2$ and $QM = QR/2 \implies \frac{AB}{PQ} = \frac{BD}{QM}$.
In $\Delta ABD$ and $\Delta PQM$, $\frac{AB}{PQ} = \frac{BD}{QM}$ and $\angle B = \angle Q \implies \Delta ABD \sim \Delta PQM$ (SAS).
Therefore, $\mathbf{\frac{AB}{PQ} = \frac{AD}{PM}}$. ■
Topic 8: High Order Thinking Skills (HOTS) & CBSE Board 5-Markers
Problem 8.1 (Square Inscribed in Right Triangle) 5 MARKS / NCERT EXEMPLAR / RD SHARMA
A square $DEFG$ is inscribed in a right-angled triangle $ABC$, right-angled at $A$. The side $DE$ lies on the hypotenuse $BC$. Prove that:
(i) $\Delta AGF \sim \Delta DBG$, (ii) $\Delta AGF \sim \Delta EFC$, (iii) $\mathbf{DE^2 = BD \times EC}$.
Step-by-Step Solution:
(i) Since $DEFG$ is a square, $GD \perp BC, FE \perp BC$, and $GF \parallel BC$.
$\angle BGD = 90^\circ - \angle B = \angle C = \angle AFG \implies \mathbf{\Delta AGF \sim \Delta DBG}$ (AA).
(ii) Similarly, $\angle CFE = 90^\circ - \angle C = \angle B = \angle AGF \implies \mathbf{\Delta AGF \sim \Delta EFC}$ (AA).
(iii) From (i) and (ii), $\Delta DBG \sim \Delta EFC \implies \frac{BD}{EF} = \frac{GD}{EC}$.
Since $GD = EF = DE$ (sides of square), we get:
$$\frac{BD}{DE} = \frac{DE}{EC} \implies \mathbf{DE^2 = BD \times EC} \quad \blacksquare$$
In the figure, $AB \perp BF, CD \perp BF$, and $EF \perp BF$. If $AB = x, CD = z$, and $EF = y$, prove that: $$\frac{1}{x} + \frac{1}{y} = \frac{1}{z}$$
Step-by-Step Solution:
In $\Delta ABF$, $CD \parallel AB \implies \frac{z}{x} = \frac{DF}{BF}$ … (I).
In $\Delta EBF$, $CD \parallel EF \implies \frac{z}{y} = \frac{BD}{BF}$ … (II).
Adding (I) and (II): $\frac{z}{x} + \frac{z}{y} = \frac{DF + BD}{BF} = \frac{BF}{BF} = 1 \implies \mathbf{\frac{1}{x} + \frac{1}{y} = \frac{1}{z}}$. ■
$\frac{1}{x} + \frac{1}{y} = \frac{1}{z}$
Problem 8.3 (Parallelogram Midpoint Proof: EL = 2BL) CBSE 2020 / RD SHARMA
Through the midpoint $M$ of side $CD$ of parallelogram $ABCD$, line $BM$ is drawn intersecting diagonal $AC$ in $L$ and $AD$ produced in $E$. Prove that $EL = 2BL$.
Step-by-Step Solution:
$\Delta BMC \cong \Delta EMD \implies BC = ED$.
$AE = AD + DE = BC + BC = 2BC$.
$\Delta AEL \sim \Delta CBL \implies \frac{EL}{BL} = \frac{AE}{BC} = \frac{2BC}{BC} = 2 \implies \mathbf{EL = 2BL}$. ■
VARDAAN COMET