Similar Figures and Basic Proportionality Theorem for CBSE Class 10

In Class 9, you studied congruence of triangles—geometric figures that have exactly the same shape and the same size. But look around you: a photograph and its enlargement, a scale model of a skyscraper, or maps of a city preserve identical geometric shapes while differing in size. Such figures are called similar figures.

In CBSE Class 10 Mathematics, Chapter 6 (Triangles) shifts focus from congruence to similarity. At the very core of triangle similarity lies one of the oldest and most celebrated theorems in geometry: the Basic Proportionality Theorem (BPT), originally established by the Greek mathematician Thales of Miletus.

What You Will Learn

• Concept of similarity: Congruence vs. Similarity of geometric figures
• Formal mathematical criteria for two polygons to be similar
• Statement and rigorous step-by-step geometric proof of the Basic Proportionality Theorem (Thales' Theorem)
• Crucial corollaries of the Basic Proportionality Theorem
• Statement and proof outline of the Converse of BPT
• High-yield CBSE board examination riders and numerical problems
• Common geometrical construction errors and exam presentation tips

1. Congruence vs. Similarity of Geometric Figures

Remember: Any two circles are always similar. Any two squares are always similar. Any two equilateral triangles are always similar.

Formal Definition of Similar Polygons:

Two polygons having the same number of sides are similar if and only if:

1. Their corresponding angles are equal, AND
2. Their corresponding sides are in the same ratio (proportional).

2. The Basic Proportionality Theorem (Thales' Theorem)

Theorem Statement (CBSE Theorem 6.1)

If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, then the other two sides are divided in the same ratio.

                                      A
                                     /                                     /                                      /  D--E                                   /  /     \                                  /  /       \                                 B--------------C

Given:

A triangle $ABC$ in which a line parallel to side $BC$ intersects side $AB$ at point $D$ and side $AC$ at point $E$ (so $DE \parallel BC$).

To Prove:

$\frac{AD}{DB} = \frac{AE}{EC}$

Construction:

1. Join vertex $B$ to $E$ and vertex $C$ to $D$.
2. Draw $DM \perp AC$ and $EN \perp AB$.

Step-by-Step Geometric Proof:

Recall that: $\text{Area of a triangle} = \frac{1}{2} \times \text{Base} \times \text{Height}$.

1. Calculate Area of $\Delta ADE$ with base $AD$: Since $EN \perp AB$, $EN$ is the altitude to base $AD$: $\text{Area}(\Delta ADE) = \frac{1}{2} \times AD \times EN$
2. Calculate Area of $\Delta BDE$ with base $DB$: Since $\Delta BDE$ is an obtuse-angled triangle, altitude $EN$ lies outside the triangle on the extended base $AB$: $\text{Area}(\Delta BDE) = \frac{1}{2} \times DB \times EN$
3. Divide the two areas: $\frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta BDE)} = \frac{\frac{1}{2} \times AD \times EN}{\frac{1}{2} \times DB \times EN} = \frac{AD}{DB} \quad \text{--- (1)}$
4. Calculate Area of $\Delta ADE$ with base $AE$: Taking $AE$ as base, $DM$ is the altitude ($DM \perp AC$): $\text{Area}(\Delta ADE) = \frac{1}{2} \times AE \times DM$
5. Calculate Area of $\Delta CDE$ with base $EC$: $\text{Area}(\Delta CDE) = \frac{1}{2} \times EC \times DM$
6. Divide these two areas: $\frac{\text{Area}(\Delta ADE)}{\text{Area}(\Delta CDE)} = \frac{\frac{1}{2} \times AE \times DM}{\frac{1}{2} \times EC \times DM} = \frac{AE}{EC} \quad \text{--- (2)}$
7. The Key Geometric Equivalence: Notice that $\Delta BDE$ and $\Delta CDE$ are on the same base $DE$ and lie between the same parallel lines $DE$ and $BC$. From Class 9 geometry, triangles on the same base and between the same parallels are equal in area: $\text{Area}(\Delta BDE) = \text{Area}(\Delta CDE) \quad \text{--- (3)}$
8. Conclusion: Comparing Equations (1), (2), and (3), the left-hand sides are equal. Therefore, their right-hand sides must be equal: $\mathbf{\frac{AD}{DB} = \frac{AE}{EC}}$ Hence, proved.

3. Important Corollaries of BPT

By adding $1$ to both sides of $\frac{AD}{DB} = \frac{AE}{EC}$: $\frac{AD}{DB} + 1 = \frac{AE}{EC} + 1 \implies \frac{AD + DB}{DB} = \frac{AE + EC}{EC} \implies \mathbf{\frac{AB}{DB} = \frac{AC}{EC}}$

Similarly, by inverting the ratio first and then adding $1$: $\mathbf{\frac{AB}{AD} = \frac{AC}{AE}}$

Important: <u>In board calculations, you can directly use the whole-side over segment ratio: $\frac{AD}{AB} = \frac{AE}{AC}$.</u>

4. Converse of Basic Proportionality Theorem

Theorem Statement (CBSE Theorem 6.2)

If a line divides any two sides of a triangle in the same ratio, then the line must be parallel to the third side.

If in $\Delta ABC$, a line $DE$ satisfies $\frac{AD}{DB} = \frac{AE}{EC}$, then $DE \parallel BC$.

5. Solved CBSE Board Examination Problems

Solved Example 1: Direct Algebraic Application

Problem: In $\Delta ABC$, $DE \parallel BC$. If $AD = x$, $DB = x - 2$, $AE = x + 2$, and $EC = x - 1$, find the value of $x$.

Solution:

1. Since $DE \parallel BC$, by the Basic Proportionality Theorem: $\frac{AD}{DB} = \frac{AE}{EC}$
2. Substitute the algebraic expressions: $\frac{x}{x - 2} = \frac{x + 2}{x - 1}$
3. Cross-multiply: $x(x - 1) = (x + 2)(x - 2)$ $x^2 - x = x^2 - 4$
4. Subtract $x^2$ from both sides: $-x = -4 \implies x = 4$
5. Therefore, <u>$x = 4$</u>.

Solved Example 2: Trapezoid Diagonals Rider (Board Classic)

Problem: $ABCD$ is a trapezium in which $AB \parallel DC$ and its diagonals intersect each other at point $O$. Show that $\frac{AO}{BO} = \frac{CO}{DO}$.

Solution:

1. Construction: Through $O$, draw a line $OE \parallel AB$ meeting side $AD$ at point $E$. Since $AB \parallel DC$, we have $OE \parallel DC$ as well.
2. In $\Delta ADC$: Since $OE \parallel DC$, by BPT: $\frac{AE}{ED} = \frac{AO}{OC} \quad \text{--- (1)}$
3. In $\Delta DAB$: Since $EO \parallel AB$, by BPT: $\frac{ED}{AE} = \frac{DO}{OB} \implies \frac{AE}{ED} = \frac{OB}{OD} = \frac{BO}{DO} \quad \text{--- (2)}$
4. Compare (1) and (2): $\frac{AO}{OC} = \frac{BO}{DO}$
5. Rearranging terms: $\mathbf{\frac{AO}{BO} = \frac{CO}{DO}}$ Hence, proved.

6. Summary and Examination Tips

Exam Tip: In the formal proof of BPT, explicitly state that "triangles on the same base and between the same parallel lines are equal in area". Omitting this geometrical justification costs 1 full mark in Section D!

Common Mistake: Forgetting that altitude $EN$ is common to both acute $\Delta ADE$ and obtuse $\Delta BDE$. An altitude can lie outside the triangle when an interior angle is obtuse.