CBSE Class 10 Mathematics Triangles Notes Set 07

Introduction to Similar Triangles and Thales Theorem

Congruent Figures

Two geometric figures which have the same shape and size are known as congruent figures. Congruent figures are alike in every respect.

Similar Figures

Geometric figures which have the same shape but different sizes are known as similar figures. Two congruent figures are always similar but two similar figures need not be congruent.

Examples: (i) any two line segments are similar; (ii) any two equiangular triangles are similar; (iii) any two squares are similar; (iv) any two circles are similar.

Similar Polygons

Two polygons are said to be similar if (i) their corresponding angles are equal and (ii) the lengths of their corresponding sides are proportional. If two polygons ABCDE and PQRST are similar we write ABCDE ~ PQRST, where the symbol '~' stands for 'is similar to'.

Equiangular Triangles

Two triangles are said to be equiangular if their corresponding angles are equal.

Similar Triangles

Two triangles ABC and DEF are said to be similar if their (i) corresponding angles are equal, that is \( \angle A = \angle D, \ \angle B = \angle E, \ \angle C = \angle F \), and (ii) corresponding sides are proportional, that is \( \dfrac{AB}{DE} = \dfrac{BC}{EF} = \dfrac{AC}{DF} \).

Basic Proportionality Theorem (BPT) or Thales Theorem

Statement. 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.

Given. A \( \triangle ABC \) in which a line parallel to side BC intersects the other two sides AB and AC at D and E respectively.
To prove. \( \dfrac{AD}{DB} = \dfrac{AE}{EC} \).
Construction. Join BE and CD and draw \( DM \perp AC \) and \( EN \perp AB \).

Proof. Area of \( \triangle ADE = \tfrac{1}{2} \times \text{base} \times \text{height} = \tfrac{1}{2} AD \times EN \). The area of \( \triangle ADE \) is written ar(ADE). So \( ar(ADE) = \tfrac{1}{2} AD \times EN \) and \( ar(BDE) = \tfrac{1}{2} DB \times EN \).

\[ \frac{ar(ADE)}{ar(BDE)} = \frac{\tfrac{1}{2} AD \times EN}{\tfrac{1}{2} DB \times EN} = \frac{AD}{DB} \qquad \text{... (i)} \]

Similarly \( ar(ADE) = \tfrac{1}{2} AE \times DM \) and \( ar(DEC) = \tfrac{1}{2} EC \times DM \), so

\[ \frac{ar(ADE)}{ar(DEC)} = \frac{AE}{EC} \qquad \text{... (ii)} \]

Note that \( \triangle BDE \) and \( \triangle DEC \) are on the same base DE and between the same parallel lines BC and DE. So \( ar(BDE) = ar(DEC) \) ... (iii). Therefore, from (i), (ii) and (iii), \( \dfrac{AD}{DB} = \dfrac{AE}{EC} \). Hence proved.

Corollary. If in a \( \triangle ABC \) a line \( DE \parallel BC \) intersects AB in D and AC in E, then (i) \( \dfrac{DB}{AD} = \dfrac{EC}{AE} \), (ii) \( \dfrac{AB}{AD} = \dfrac{AC}{AE} \), (iii) \( \dfrac{AD}{AB} = \dfrac{AE}{AC} \), (iv) \( \dfrac{AB}{DB} = \dfrac{AC}{EC} \), (v) \( \dfrac{DB}{AB} = \dfrac{EC}{AC} \).

Converse of the Basic Proportionality Theorem

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

Given. A \( \triangle ABC \) and a line intersecting AB at D and AC at E such that \( \dfrac{AD}{DB} = \dfrac{AE}{EC} \).
To prove. \( DE \parallel BC \).

Proof. If possible, let DE not be parallel to BC. Then there must be another line through D, which is parallel to BC. Let \( DF \parallel BC \). By Thales' theorem \( \dfrac{AD}{DB} = \dfrac{AF}{FC} \) ... (i). But \( \dfrac{AD}{DB} = \dfrac{AE}{EC} \) (given) ... (ii). From (i) and (ii), \( \dfrac{AF}{FC} = \dfrac{AE}{EC} \). Adding 1 to both sides, \( \dfrac{AF + FC}{FC} = \dfrac{AE + EC}{EC} \), that is \( \dfrac{AC}{FC} = \dfrac{AC}{EC} \), so FC = EC. This is possible only when E and F coincide. Hence \( DE \parallel BC \).

Some Important Results and Theorems

1. The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle (called the Angle Bisector Theorem).

2. In a triangle ABC, if D is a point on BC such that D divides BC in the ratio AB : AC, then AD is the bisector of \( \angle A \).

3. The external bisector of an angle of a triangle divides the opposite side externally in the ratio of the sides containing the angle.

4. The line drawn from the mid-point of one side of a triangle parallel to another side bisects the third side.

5. The line joining the mid-points of two sides of a triangle is parallel to the third side.

6. The diagonals of a trapezium divide each other proportionally.

7. If the diagonals of a quadrilateral divide each other proportionally, then it is a trapezium.

8. Any line parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally.

9. If three or more parallel lines are intersected by two transversals, then the intercepts made by them on the transversals are proportional.

Solved Examples

Check Your Progress

1. If three or more parallel lines are intersected by two transversals, prove that the intercepts made by them on the transversals are proportional.
Answer:
Let the parallel lines l, m, n meet one transversal at A, B, C and the other at D, E, F. Join AF, meeting m at G.
In \( \triangle ACF \), \( BG \parallel CF \), so \( \dfrac{AB}{BC} = \dfrac{AG}{GF} \). In \( \triangle FAD \), \( GE \parallel AD \), so \( \dfrac{AG}{GF} = \dfrac{DE}{EF} \).
Hence \( \dfrac{AB}{BC} = \dfrac{DE}{EF} \).

2. Using the converse of BPT, prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side.
Answer:
In \( \triangle ABC \) let D and E be the midpoints of AB and AC. Then AD = DB and AE = EC, so \( \dfrac{AD}{DB} = 1 = \dfrac{AE}{EC} \).
By the converse of the Basic Proportionality Theorem, \( DE \parallel BC \).

3. If D, E are points on the sides AB and AC of \( \triangle ABC \) such that AD = 6 cm, BD = 9 cm, AE = 8 cm and EC = 12 cm, prove that \( DE \parallel BC \).
Answer:
\( \dfrac{AD}{DB} = \dfrac{6}{9} = \dfrac{2}{3} \) and \( \dfrac{AE}{EC} = \dfrac{8}{12} = \dfrac{2}{3} \). The ratios are equal, so by the converse of BPT, \( DE \parallel BC \).

5. ABCD is a trapezium in which AB is parallel to DC. If the diagonals intersect at O, prove that \( AO \cdot DO = BO \cdot CO \).
Answer:
The diagonals of a trapezium divide each other proportionally, so \( \dfrac{AO}{OC} = \dfrac{BO}{OD} \). Cross-multiplying, \( AO \cdot OD = BO \cdot OC \), that is \( AO \cdot DO = BO \cdot CO \).