Chapter 15: Similarity (With Applications to Maps & Models) - Ultimate 100/100 Master Study Guide

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Topic 1: Introduction to Similarity & Comparison with Congruence

Similarity vs Congruency

Similarity of Figures ($\sim$): Two figures are similar if they have the same shape, but may differ in size.
• Same shape: Angles of one figure are equal to corresponding angles of the other figure.
• Same size: Sides of one figure are equal to corresponding sides of the other figure.
Congruency of Figures ($\cong$): Two figures are congruent if they have both the same shape and the same size.
Fundamental Axiom: Congruent figures are ALWAYS similar, whereas similar figures are NOT necessarily congruent.

Feature	Congruent Triangles ($\Delta ABC \cong \Delta DEF$)	Similar Triangles ($\Delta ABC \sim \Delta DEF$)
Shape	Identical	Identical
Size	Exactly Equal ($k = 1$)	Scaled by factor $k > 0$
Corresponding Angles	Strictly Equal ($\angle A = \angle D, \angle B = \angle E, \angle C = \angle F$)	Strictly Equal ($\angle A = \angle D, \angle B = \angle E, \angle C = \angle F$)
Corresponding Sides	Strictly Equal ($AB = DE, BC = EF, AC = DF$)	Proportional: $\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = k$
Core Keyword	Equality	Proportionality

ICSE True / False Exam Checklist

Two similar polygons are NOT necessarily congruent. (True)
Two congruent polygons are necessarily similar. (True)
All equiangular triangles are similar. (True — AAA postulate)
All isosceles triangles are similar. (False — angle between equal sides may differ)
All isosceles-right triangles are similar. (True — angles are always $45^\circ, 45^\circ, 90^\circ$)
The diagonals of a trapezium divide each other proportionally. (True)

Topic 2: Corresponding Sides & Vertex Sequencing Rules

Corresponding Parts Identification Rules

Rule 1 (Sides opposite to equal angles): In similar triangles, the sides opposite to equal angles are called corresponding sides and are in proportion.
If $\angle A = \angle P$, then side $BC$ corresponds to side $QR$.
If $\angle B = \angle R$, then side $AC$ corresponds to side $PQ$.
If $\angle C = \angle Q$, then side $AB$ corresponds to side $PR$. $$\implies \mathbf{\frac{BC}{QR} = \frac{AC}{PQ} = \frac{AB}{PR} \iff \Delta ABC \sim \Delta PRQ}$$
Rule 2 (Angles opposite to proportional sides): In similar triangles, the angles opposite to proportional sides are equal.
If $\frac{AB}{EF} = \frac{BC}{DF} = \frac{AC}{DE} \implies \angle C = \angle D, \angle A = \angle E, \angle B = \angle F \implies \mathbf{\Delta ABC \sim \Delta EFD}$.
Rule 3 (Order of Vertices Strictness): The names of two similar triangles must be written in one-to-one vertex correspondence ($A \leftrightarrow D, B \leftrightarrow E, C \leftrightarrow F \implies \Delta ABC \sim \Delta DEF$). Writing $\Delta ABC \sim \Delta DFE$ is strictly penalised in board evaluation.

Fig 15.1: Vertex Correspondence $A \leftrightarrow D, B \leftrightarrow E, C \leftrightarrow F \implies \Delta ABC \sim \Delta DEF$

Topic 3: Conditions of Similar Triangles (SAS, AA, SSS) & Right-Triangle Altitudes

The 3 Similarity Criteria

1. SAS Postulate: If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional: $$\text{If } \angle A = \angle D \text{ and } \frac{AB}{DE} = \frac{AC}{DF} \implies \mathbf{\Delta ABC \sim \Delta DEF}$$
2. AA (or AAA) Postulate: If two angles of one triangle are respectively equal to two angles of another triangle: $$\text{If } \angle A = \angle D \text{ and } \angle B = \angle E \implies \mathbf{\Delta ABC \sim \Delta DEF}$$ (Since angle sum is $180^\circ$, third angles are automatically equal).
3. SSS Postulate: If all three corresponding sides of two triangles are in the same ratio: $$\text{If } \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} \implies \mathbf{\Delta ABC \sim \Delta PQR}$$

Theorem: Altitude Drawn to Hypotenuse of a Right-Angled Triangle

Statement: A perpendicular drawn from the vertex of the right angle of a right-angled triangle to its hypotenuse divides the triangle into two triangles which are similar to each other and to the original whole triangle.

Fig 15.2: In right $\Delta ABC$ ($\angle A = 90^\circ$), $AD \perp BC \implies \Delta DBA \sim \Delta DAC \sim \Delta ABC$

The 3 Fundamental Geometric Formulas:

$\Delta DBA \sim \Delta CBA \implies \frac{AB}{BC} = \frac{BD}{AB} \implies \mathbf{AB^2 = BD \times BC}$
$\Delta DAC \sim \Delta ABC \implies \frac{AC}{BC} = \frac{DC}{AC} \implies \mathbf{AC^2 = CD \times BC}$
$\Delta DBA \sim \Delta DAC \implies \frac{AD}{CD} = \frac{BD}{AD} \implies \mathbf{AD^2 = BD \times CD}$ (Geometric Mean Formula)
$\mathbf{\frac{AB^2}{AC^2} = \frac{BD \times BC}{CD \times BC} = \frac{BD}{CD}}$

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Exercise 15(A) Practice Kit: Core Similarity & Altitudes

Problem 3.1 (Right Triangle Altitude Calculation) TEXTBOOK EX 15(A) Q19

In $\Delta ABC$, $\angle B = 90^\circ$ and $BD \perp AC$. If $CD = 10\text{ cm}$ and $BD = 8\text{ cm}$, find $AD$. If $AC = 18\text{ cm}$ and $AD = 6\text{ cm}$, find $BD$.

Step-by-Step Solution:

(i) $BD^2 = AD \times CD \implies 8^2 = AD \times 10 \implies 64 = 10 AD \implies \mathbf{AD = 6.4\text{ cm}}$.
(ii) $CD = AC - AD = 18 - 6 = 12\text{ cm}$. $$BD^2 = AD \times CD = 6 \times 12 = 72 \implies \mathbf{BD = \sqrt{72} = 6\sqrt{2}\text{ cm} \approx 8.48\text{ cm}}$$

Problem 3.2 (Three Concurrent Lines Intercepting Parallels) TEXTBOOK EX 15(A) Q2

Lines $l$ and $m$ are parallel. Three concurrent lines through point $O$ meet line $l$ at $A, B, C$ and line $m$ at $P, Q, R$. Prove that: $$\frac{AB}{BC} = \frac{QR}{PQ} \quad \left(\text{or } \frac{AB}{BC} = \frac{PQ}{QR}\right)$$

Step-by-Step Solution:

$\Delta OAB \sim \Delta OQR \implies \frac{AB}{QR} = \frac{OB}{OQ}$.
$\Delta OBC \sim \Delta OPQ \implies \frac{BC}{PQ} = \frac{OB}{OQ}$.
Equating: $\frac{AB}{QR} = \frac{BC}{PQ} \implies \mathbf{\frac{AB}{BC} = \frac{QR}{PQ}}$. ■

Problem 3.3 (Rhombus Diagonal Intersection Proof) TEXTBOOK EX 15(A) Q16

$ABCD$ is a rhombus. $DPR$ and $CBR$ are straight lines intersecting at $R$, and diagonal $AC$ intersects $DR$ at $P$. Prove that $DP \times CR = DC \times PR$.

Step-by-Step Solution:

In $\Delta RDC$, since $AB \parallel DC \implies PB \parallel DC$ in line segment:
$\Delta RPB \sim \Delta RDC \implies \frac{PR}{DR} = \frac{PB}{DC}$ or in $\Delta DPC$ and $\Delta RPA$: $\frac{DP}{PR} = \frac{DC}{CR} \implies \mathbf{DP \times CR = DC \times PR}$. ■

Topic 4: Basic Proportionality Theorem (Thales Theorem) with Applications

Theorem 1: Basic Proportionality Theorem (BPT)

Statement: A line drawn parallel to one side of a triangle divides the other two sides proportionally.

$$\text{If } DE \parallel BC \implies \mathbf{\frac{AD}{DB} = \frac{AE}{EC}} \quad \text{and} \quad \mathbf{\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}}$$

Fig 15.3: BPT Parallel Line Proportionality $DE \parallel BC$

Formal Proof (By Similarity):

In $\Delta ADE$ and $\Delta ABC$: $\angle ADE = \angle ABC$ (Corresponding angles), $\angle AED = \angle ACB$ (Corresponding angles), and $\angle A = \angle A$ (Common).
$\therefore \Delta ADE \sim \Delta ABC$ (by AAA Postulate).
$\implies \frac{AB}{AD} = \frac{AC}{AE} \implies \frac{AD + DB}{AD} = \frac{AE + EC}{AE} \implies 1 + \frac{DB}{AD} = 1 + \frac{EC}{AE}$.
Subtracting 1 from both sides: $\frac{DB}{AD} = \frac{EC}{AE} \implies \mathbf{\frac{AD}{DB} = \frac{AE}{EC}}$. ■

Converse of BPT & Three Parallel Lines

Converse of BPT: If a line divides any two sides of a triangle in the same ratio ($\frac{AD}{DB} = \frac{AE}{EC}$), then the line is parallel to the third side ($DE \parallel BC$).
Three Parallel Lines Theorem: If three parallel lines $l \parallel m \parallel n$ are intersected by two transversals at $A, B, C$ and $P, Q, R$, then: $$\mathbf{\frac{AB}{BC} = \frac{PQ}{QR}}$$

Exercise 15(B) Practice Kit: BPT Applications

Problem 4.1 (Three Parallel Lines in Trapezium) TEXTBOOK EX 15(B) Q7

In the figure, $AB, CD$ and $EF$ are parallel lines. Given $AB = 7.5\text{ cm}, DC = y\text{ cm}, EF = 4.5\text{ cm}, BC = x\text{ cm}$, and $CE = 3\text{ cm}$, calculate the values of $x$ and $y$.

Step-by-Step Solution:

In $\Delta ABE$, $DC \parallel AB \implies \frac{CE}{BE} = \frac{DC}{AB} \implies \frac{3}{x + 3} = \frac{y}{7.5}$.
Also in $\Delta BEF$, $DC \parallel EF \implies \frac{BC}{BE} = \frac{DC}{EF} \implies \frac{x}{x + 3} = \frac{y}{4.5}$.
Dividing: $\frac{3}{x} = \frac{4.5}{7.5} = \frac{3}{5} \implies \mathbf{x = 5\text{ cm}}$.
Substitute $x = 5$: $\frac{3}{5 + 3} = \frac{y}{7.5} \implies \frac{3}{8} = \frac{y}{7.5} \implies y = \frac{22.5}{8} = \mathbf{2.8125\text{ cm}}$.

Problem 4.2 (Nested Parallel Segments Proof) TEXTBOOK EX 15(B) Q9

In $\Delta ABC$, $M$ is a point on $BC$. Lines $DM \parallel AC$ and $EM \parallel AB$ meet $AB$ and $AC$ at $D$ and $E$. Prove that $\frac{BD}{DA} = \frac{BM}{MC}$ and $\frac{CE}{EA} = \frac{CM}{MB}$.

Step-by-Step Solution:

In $\Delta BAC$, $DM \parallel AC \implies \frac{BD}{DA} = \frac{BM}{MC}$ (by BPT).
In $\Delta CAB$, $EM \parallel AB \implies \frac{CE}{EA} = \frac{CM}{MB}$ (by BPT).
Multiplying both ratios: $\frac{BD}{DA} \times \frac{CE}{EA} = \frac{BM}{MC} \times \frac{MC}{BM} = \mathbf{1}$. ■

Topic 5: Relation Between Areas of Two Similar Triangles

Theorem 2: Ratio of Areas of Similar Triangles

Statement: The areas of two similar triangles are proportional to the squares on their corresponding sides (and altitudes, medians, and perimeters).

$$\mathbf{\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta DEF)} = \frac{AB^2}{DE^2} = \frac{BC^2}{EF^2} = \frac{AC^2}{DF^2} = \frac{AM^2}{DN^2} = \left(\frac{\text{Perimeter}_1}{\text{Perimeter}_2}\right)^2 = k^2}$$

Fig 15.4: Area Ratio $= \frac{BC \times AM}{EF \times DN} = \frac{BC}{EF} \times \frac{BC}{EF} = \frac{BC^2}{EF^2}$

Formal Statement-Reason Proof:

$\text{Area}(\Delta ABC) = \frac{1}{2} BC \times AM$ and $\text{Area}(\Delta DEF) = \frac{1}{2} EF \times DN \implies \frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta DEF)} = \frac{BC}{EF} \times \frac{AM}{DN}$.
In $\Delta ABM$ and $\Delta DEN$: $\angle B = \angle E$ and $\angle AMB = \angle DNE = 90^\circ \implies \Delta ABM \sim \Delta DEN \implies \frac{AM}{DN} = \frac{AB}{DE}$.
Since $\Delta ABC \sim \Delta DEF \implies \frac{AB}{DE} = \frac{BC}{EF}$. Therefore $\frac{AM}{DN} = \frac{BC}{EF}$.
Substituting: $\mathbf{\frac{\text{Area}(\Delta ABC)}{\text{Area}(\Delta DEF)} = \frac{BC}{EF} \times \frac{BC}{EF} = \frac{BC^2}{EF^2}}$. ■

CRITICAL EXAM DISTINCTION (SQUARED VS LINEAR AREA RATIO)

Similar Triangles ($\Delta ADE \sim \Delta ABC$): Areas are proportional to the SQUARE of sides: $$\frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta ABC)} = \left(\frac{DE}{BC}\right)^2 = \left(\frac{AD}{AB}\right)^2$$
Triangles with Common Vertex & Collinear Bases: Areas are directly proportional to the LINEAR ratio of bases: $$\frac{\text{Area}(\Delta ABE)}{\text{Area}(\Delta ACE)} = \frac{BE}{CE}$$
Median Area Rule: A median divides a triangle into two triangles of strictly equal area ($\text{Area}(\Delta ABD) = \text{Area}(\Delta ACD) = \frac{1}{2}\text{Area}(\Delta ABC)$).
Trapezium Side-Triangles Rule: In trapezium $ABCD$ ($AB \parallel DC$) with diagonals intersecting at $P$: $\mathbf{\text{Area}(\Delta APD) = \text{Area}(\Delta BPC)}$.

Exercise 15(C) Practice Kit: Area Theorems

Problem 5.1 (Line Bisecting Triangle Area) TEXTBOOK EX 15(C) Q5

$ABC$ is a triangle. $PQ$ is a line segment intersecting $AB$ in $P$ and $AC$ in $Q$ such that $PQ \parallel BC$ and divides $\Delta ABC$ into two parts equal in area. Find the ratio $\frac{BP}{AB}$.

Step-by-Step Solution:

$\text{Area}(\Delta APQ) = \frac{1}{2}\text{Area}(\Delta ABC) \implies \frac{\text{Area}(\Delta APQ)}{\text{Area}(\Delta ABC)} = \frac{1}{2}$.
Since $\Delta APQ \sim \Delta ABC \implies \left(\frac{AP}{AB}\right)^2 = \frac{1}{2} \implies \frac{AP}{AB} = \frac{1}{\sqrt{2}}$. $$\frac{BP}{AB} = \frac{AB - AP}{AB} = 1 - \frac{AP}{AB} = 1 - \frac{1}{\sqrt{2}} = \mathbf{\frac{\sqrt{2} - 1}{\sqrt{2}} = \frac{2 - \sqrt{2}}{2}}$$

Problem 5.2 (Parallelogram & Extended Line Area) TEXTBOOK EX 15(C) Q8

$ABCD$ is a parallelogram. $P$ is a point on $BC$ such that $BP : PC = 1 : 2$. $DP$ produced meets $AB$ produced at $Q$. Given the area of $\Delta CPQ = 20\text{ cm}^2$, find: (i) Area of $\Delta CDP$, (ii) Area of parallelogram $ABCD$.

Step-by-Step Solution:

$\Delta QBP \sim \Delta DCP \implies \frac{BP}{PC} = \frac{1}{2} \implies \frac{\text{Area}(\Delta QBP)}{\text{Area}(\Delta DCP)} = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$.
In $\Delta QDC$, $P$ is on $DC$ transversal $\implies \text{Area}(\Delta CDP) = \mathbf{40\text{ cm}^2}$.
$\text{Area}(\text{Parallelogram } ABCD) = 2 \times \text{Area}(\Delta BCD) = 2 \times (40 + 20) = \mathbf{120\text{ cm}^2}$.

Topic 6: Similarity as a Size Transformation & Applications to Maps and Models

Size Transformation (Enlargement & Reduction)

If a triangle $ABC$ is enlarged/reduced about a fixed point $P$ by scale factor $k$ ($PA' = k \cdot PA, PB' = k \cdot PB, PC' = k \cdot PC$):

$k > 1 \implies$ Enlargement: Image is larger than object.
$k < 1 \implies$ Reduction: Image is smaller than object.
$k = 1 \implies$ Identity Transformation: Image is congruent to object.

Master Scale Relations for Maps & Scale Models:

If the scale factor of a map/model is $k = \frac{\text{Model Length}}{\text{Actual Length}} = \frac{1}{n}$:

Dimension	Transformation Relation	Scale Ratio Formula
1D: Length / Height / Perimeter	$\text{Length}_{\text{model}} = k \times \text{Length}_{\text{actual}}$	$\mathbf{\frac{\text{Length}_{\text{model}}}{\text{Length}_{\text{actual}}} = k = \frac{1}{n}}$
2D: Surface Area / Plot Area	$\text{Area}_{\text{model}} = k^2 \times \text{Area}_{\text{actual}}$	$\mathbf{\frac{\text{Area}_{\text{model}}}{\text{Area}_{\text{actual}}} = k^2 = \frac{1}{n^2}}$
3D: Volume / Capacity / Mass	$\text{Volume}_{\text{model}} = k^3 \times \text{Volume}_{\text{actual}}$	$\mathbf{\frac{\text{Volume}_{\text{model}}}{\text{Volume}_{\text{actual}}} = k^3 = \frac{1}{n^3}}$

Essential Unit Conversion Matrix

$1\text{ km} = 1000\text{ m} = 100,000\text{ cm} = \mathbf{10^5\text{ cm}}$
$1\text{ km}^2 = (10^5\text{ cm})^2 = \mathbf{10^{10}\text{ cm}^2} = \mathbf{10^6\text{ m}^2}$
$1\text{ m}^2 = (100\text{ cm})^2 = \mathbf{10^4\text{ cm}^2} = 10,000\text{ cm}^2$
$1\text{ m}^3 = (100\text{ cm})^3 = \mathbf{10^6\text{ cm}^3} = \mathbf{1000\text{ litres}}$
$1\text{ litre} = \mathbf{1000\text{ cm}^3}$

Exercise 15(D) Practice Kit: Maps & Scale Models

Problem 6.1 (Map Scale & Plot Area) TEXTBOOK EX 15(D) Q6

On a map drawn to a scale of $1 : 250,000$, a triangular plot of land has measurements $AB = 3\text{ cm}, BC = 4\text{ cm}, \angle B = 90^\circ$. Calculate:
(i) The actual lengths of $AB$ and $BC$ in km.
(ii) The actual area of the plot in sq. km.

Step-by-Step Solution:

$k = \frac{1}{250,000}$.
(i) $\text{Actual } AB = 3 \times 250,000\text{ cm} = 750,000\text{ cm} = \frac{750,000}{100,000} = \mathbf{7.5\text{ km}}$.
$\text{Actual } BC = 4 \times 250,000\text{ cm} = 1,000,000\text{ cm} = \mathbf{10.0\text{ km}}$.
(ii) $\text{Area on map} = \frac{1}{2} \times 3 \times 4 = 6\text{ cm}^2$. $$\text{Actual Area} = 6 \times (250,000)^2\text{ cm}^2 = 6 \times 6.25 \times 10^{10}\text{ cm}^2 = 3.75 \times 10^{11}\text{ cm}^2$$ $$\text{Actual Area in km}^2 = \frac{3.75 \times 10^{11}}{10^{10}} = \mathbf{37.5\text{ sq. km}}$$

Problem 6.2 (Ship Model Dimensions, Area & Volume) TEXTBOOK EX 15(D) Q7

A model of a ship is made to a scale of $1 : 200$.
(i) If the length of the model is $4\text{ m}$, calculate the length of the ship.
(ii) The area of the deck of the ship is $160,000\text{ m}^2$, find the area of the deck of the model.
(iii) The volume of the model is $200\text{ litres}$; calculate the volume of the ship in $\text{m}^3$.

Step-by-Step Solution:

(i) $\text{Ship Length} = 4\text{ m} \times 200 = \mathbf{800\text{ m}}$.
(ii) $\text{Model Deck Area} = \left(\frac{1}{200}\right)^2 \times 160,000 = \frac{160,000}{40,000} = \mathbf{4\text{ m}^2}$.
(iii) $\text{Ship Volume} = 200\text{ litres} \times (200)^3 = 200 \times 8 \times 10^6 = 1.6 \times 10^9\text{ litres}$. $$\text{Ship Volume in m}^3 = \frac{1.6 \times 10^9}{1000} = \mathbf{1.6 \times 10^6\text{ m}^3} = \mathbf{1,600,000\text{ m}^3}$$

Problem 6.3 (Rectangular Tank Scale Model) TEXTBOOK EX 15 EXAMPLE 15

A rectangular tank has length $= 4\text{ m}$, width $= 3\text{ m}$, and capacity $= 30\text{ m}^3$. A small model of the tank is made with capacity $240\text{ cm}^3$. Find: (i) Dimensions of the model, (ii) Ratio between the total surface area of the tank and its model.

Step-by-Step Solution:

Height of tank $= \frac{30}{4 \times 3} = 2.5\text{ m}$. $$\text{Volume ratio } k^3 = \frac{240\text{ cm}^3}{30\text{ m}^3} = \frac{240}{30 \times 10^6\text{ cm}^3} = \frac{1}{125,000} \implies k = \frac{1}{50}$$ (i) $\text{Length}_{\text{model}} = \frac{400\text{ cm}}{50} = \mathbf{8\text{ cm}}$, $\text{Width}_{\text{model}} = \frac{300\text{ cm}}{50} = \mathbf{6\text{ cm}}$, $\text{Height}_{\text{model}} = \frac{250\text{ cm}}{50} = \mathbf{5\text{ cm}}$.
(ii) $\frac{\text{Total S.A. of tank}}{\text{Total S.A. of model}} = \frac{1}{k^2} = (50)^2 = \mathbf{2500 : 1}$.

Topic 7: Comprehensive ICSE Board Examination Repository (Exercise 15E)

Exercise 15(E) High-Yield Board Exam Problems

Problem 7.1 (Three Perpendiculars: 1/x + 1/y = 1/z) BOARD CLASSIC

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): $\frac{z}{x} + \frac{z}{y} = \frac{DF + BD}{BF} = \frac{BF}{BF} = 1 \implies \mathbf{\frac{1}{x} + \frac{1}{y} = \frac{1}{z}}$. ■

Problem 7.2 (Centroid 2:1 Ratio from Medians Similarity) TEXTBOOK EX 15(E) Q16

In $\Delta ABC$, medians $AD$ and $CE$ intersect at $G$. $DF$ is drawn parallel to $CE$. Prove that: (i) $EF = FB$, (ii) $AG : GD = 2 : 1$.

Step-by-Step Solution:

(i) In $\Delta BCE$, $D$ is mid-point of $BC$ and $DF \parallel CE \implies F$ is mid-point of $BE \implies \mathbf{EF = FB}$.
(ii) $\Delta EDG \sim \Delta BCG \implies \frac{GD}{GB} = \frac{ED}{BC} = \frac{1}{2} \implies \mathbf{AG : GD = 2 : 1}$. ■

Problem 7.3 (Board Exam 2014 Solved Proof) ICSE 2014

In $\Delta ABC$, $\angle ABC = \angle DAC$. $AB = 8\text{ cm}, AC = 4\text{ cm}, AD = 5\text{ cm}$.
(i) Prove that $\Delta ACD \sim \Delta BCA$.
(ii) Find $BC$ and $CD$.
(iii) Find $\frac{\text{Area}(\Delta ACD)}{\text{Area}(\Delta BCA)}$.

Step-by-Step Solution:

(i) $\angle CAD = \angle CBA$ (given) and $\angle C = \angle C$ (common) $\implies \mathbf{\Delta ACD \sim \Delta BCA}$ (AA).
(ii) $\frac{AC}{BC} = \frac{CD}{CA} = \frac{AD}{BA} \implies \frac{4}{BC} = \frac{CD}{4} = \frac{5}{8}$.
$BC = \frac{32}{5} = \mathbf{6.4\text{ cm}}$, $CD = \frac{20}{8} = \mathbf{2.5\text{ cm}}$.
(iii) $\frac{\text{Area}(\Delta ACD)}{\text{Area}(\Delta BCA)} = \left(\frac{AC}{BC}\right)^2 = \left(\frac{4}{6.4}\right)^2 = \left(\frac{5}{8}\right)^2 = \mathbf{\frac{25}{64}}$.

Problem 7.4 (Parallelogram Midpoint Proof: EL = 2BL) TEXTBOOK EX 15(E) Q25

Through the midpoint $M$ of side $CD$ of parallelogram $ABCD$, line $BM$ is drawn intersecting $AC$ in $L$ and $AD$ produced in $E$. Prove that $EL = 2BL$.

Step-by-Step Solution:

$\Delta BMC \cong \Delta EMD \implies BC = ED$.
$AE = AD + DE = BC + BC = 2BC$.
$\Delta AEL \sim \Delta CBL \implies \frac{EL}{BL} = \frac{AE}{BC} = \frac{2BC}{BC} = 2 \implies \mathbf{EL = 2BL}$. ■

Topic 8: 100/100 ICSE Examination Master Formula Checklist

Rapid Revision Matrix

Similar Triangles: $\frac{AB}{DE} = \frac{BC}{EF} = \frac{AC}{DF} = \frac{\text{Alt}_1}{\text{Alt}_2} = \frac{\text{Med}_1}{\text{Med}_2} = \frac{\text{Perimeter}_1}{\text{Perimeter}_2} = k$.
BPT: $DE \parallel BC \implies \frac{AD}{DB} = \frac{AE}{EC}$ and $\frac{AD}{AB} = \frac{AE}{AC} = \frac{DE}{BC}$.
Right Triangle Altitude: $AD^2 = BD \times CD$, $AB^2 = BD \times BC$, $AC^2 = CD \times BC$, $\frac{AB^2}{AC^2} = \frac{BD}{CD}$.
Area Ratio: $\frac{\text{Area}_1}{\text{Area}_2} = \left(\frac{\text{Side}_1}{\text{Side}_2}\right)^2 = \left(\frac{\text{Altitude}_1}{\text{Altitude}_2}\right)^2 = \left(\frac{\text{Median}_1}{\text{Median}_2}\right)^2 = k^2$.
Common Vertex Area Ratio: $\frac{\text{Area}(\Delta ABE)}{\text{Area}(\Delta ACE)} = \frac{BE}{CE}$ (Linear ratio).
Maps & Models: $\text{Length} \propto k, \ \text{Area} \propto k^2, \ \text{Volume} \propto k^3$.
Key Units: $1\text{ km} = 10^5\text{ cm}$, $1\text{ km}^2 = 10^{10}\text{ cm}^2 = 10^6\text{ m}^2$, $1\text{ m}^3 = 1000\text{ litres} = 10^6\text{ cm}^3$.