Basic Proportionality Theorem (Thales Theorem)

Triangles of Class 10

Basic Proportionality Theorem (Thales Theorem)

THEOREM 1:

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.

Given: A triangle ABC in which a line parallel to side BC intersects other two sides AB and AC at D and E respectively.

Basic Proportionality Theorem (Thales Theorem)

To prove: Basic Proportionality Theorem (Thales Theorem)

Construction: Join BE and CD and draw DM ⊥ AC and EN ⊥ AB.

Proof: area of Δ ADE Basic Proportionality Theorem (Thales Theorem)

(Taking AD as base)

So, Basic Proportionality Theorem (Thales Theorem) [The area of Δ ADE is denoted as ar (ADE)].

Similarly, Basic Proportionality Theorem (Thales Theorem)

Basic Proportionality Theorem (Thales Theorem)

[Δ BDE and DEC are on the same base DE and between the same parallels BC and DE.]

Therefore, from (i), (ii) and (iii), we have:

Basic Proportionality Theorem (Thales Theorem).

Corollary: From above equation we have

Basic Proportionality Theorem (Thales Theorem).

Adding '1' to both sides we have

Basic Proportionality Theorem (Thales Theorem)

THEOREM 2:

(Converse of BPT theorem) If a line divides any two sides of a triangle in the same ratio, prove that it is parallel to the third side.

Given: In ΔABC, DE is a straight line such that Basic Proportionality Theorem (Thales Theorem).

To prove: DE || BC.

Construction: If DE is not parallel to BC, draw DF meeting AC at F.

Proof: In ΔABC, let DF || BC

Basic Proportionality Theorem (Thales Theorem)

[∴ A line drawn parallel to one side of a Δ divides the other two sides in the same ratio.]

But Basic Proportionality Theorem (Thales Theorem). …(ii) [given]

From (i) and (ii), we get

Basic Proportionality Theorem (Thales Theorem).

⇒ FC = EC.

It is possible only when E and F coincide

Hence, DE || BC.

SOME IMPORTANT RESULTS AND THEOREMS:

(i) The internal bisector of an angle of a triangle divides the opposite side internally in the ratio of the sides containing the angle.

(ii) 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 ∠A.

(iii) The external bisector of an angle of a triangle divides the opposite sides externally in the ratio of the sides containing the angle.

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

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

(vi) The diagonals of a trapezium divide each other proportionally.

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

(viii) Any line parallel to the parallel sides of a trapezium divides the non-parallel sides proportionally.

(ix) If three or more parallel lines are intersected by two transversal, then the intercepts made by them on the transversal are proportional.

If any students need to take the online test to check their concepts or undertstanding then they can visit Question for Basic Proportionality Theorm.

Question 1. In a ΔABC, D and E are points on the sides AB and AC respectively such that DE || BC. If AD = 4x – 3, AE = 8x – 7, BD = 3x – 1 and CE = 5x – 3, find the value of x.

Solution: In ΔABC, we have

DE||BC.

Basic Proportionality Theorem (Thales Theorem). [By Basic Proportionality Theorem]

Basic Proportionality Theorem (Thales Theorem)

⇒ 20x2 – 15x – 12x + 9 = 24x2 – 21x – 8x + 7

⇒ 20x2 – 27x + 9 = 24x2 – 29x + 7

⇒ 4x2 – 2x – 2 = 0

⇒ 2x2 – x – 1 = 0

⇒ (2x + 1) (x – 1) = 0

⇒ x = 1 or x = –1/2

So, the required value of x is 1.

[x = - 1/2 is neglected as length cannot be negative].

Question 2. D and E are respectively the points on the sides AB and AC of a ΔABC such that AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm, show that DE || BC.

65.jpg

Solution: We have,

AB = 12 cm, AC = 18 m, AD = 8 cm and AE = 12 cm.

∴ BD = AB - AD = (12 – 8) cm = 4 cm.

CE = AC – AE = (18 12) cm = 6 cm.

Basic Proportionality Theorem (Thales Theorem).

Thus, DE divides sides AB and AC of ΔABC in the same ratio. Therefore, by the converse of basic proportionality theorem we have DE||BC.

Question 3. In a trapezium ABCD AB||DC and DC = 2AB. EF drawn parallel to AB cuts AD in F and BC in E such thatBasic Proportionality Theorem (Thales Theorem). Diagonal DB intersects EF at G. Prove that 7FE = 10AB.

66.jpg

Solution: In ΔDFG and ΔDAB,

∠1 = ∠2 [Corresponding ∠s ∴ AB || FG]

∠FDG = ∠ADB [Common]

∴ ΔDFG ~ ΔDAB [By AA rule of similarity]

Basic Proportionality Theorem (Thales Theorem)

From (i) and (ii), we get

FG/AB = 4/7 i.e. FG = 4/7 AB ......(iii)

In ΔBEG and ΔBCD, we have

∠BEG = ∠BCD [Corresponding angle ∴ EG||CD]

∠GBE = ∠DBC [Common]

∴ ΔBEG ~ ΔBCD [By AA rule of similarity]

Basic Proportionality Theorem (Thales Theorem)

Hence proved.

Question 4. In ΔABC, if AD is the bisector of ∠A, prove that Basic Proportionality Theorem (Thales Theorem).

Basic Proportionality Theorem (Thales Theorem)

Solution: In ΔABC, AD is the bisector of ∠A.

∴ AB/AC = BD/DC....(i) [By internal bisector theorem]

From A draw AL⊥BC

Basic Proportionality Theorem (Thales Theorem).

[From (i)] Hence Proved.

Question 5. ∠BAC = 90o, AD is its bisector. IF DE⊥ AC, prove that DE × (AB + AB) = AB × AC.

Solution: It is given that AD is the bisector of ∠A of ΔABC.

Basic Proportionality Theorem (Thales Theorem)

In Δ’s CDE and CBA, we have

∠DCE = ∠BCA [Common]

∠DEC = ∠BAC. [Each equal to 90o]

So by AA-criterion of similarity

Δ CDE ~ Δ CBA

Basic Proportionality Theorem (Thales Theorem)

⇒ DE × (AB + AC) = AB × AC.

Question 6. In the given figure, PA, QB and RC are each perpendicular to AC. Prove thatBasic Proportionality Theorem (Thales Theorem).

Solution: In ΔPAC, we have BQ||AP

Basic Proportionality Theorem (Thales Theorem)

In ΔACR, we have BQ||CR

Basic Proportionality Theorem (Thales Theorem)

Hence Proved

Question 7. In the given figure, AB||CD. Find the value of x.

Solution: Since the diagonals of a trapezium divide each other proportionally.

70.jpg

Basic Proportionality Theorem (Thales Theorem)

Basic Proportionality Theorem (Thales Theorem)

⇒ 12x – 76 = x2 – 4x – 3x + 12

⇒ x2 – 19x + 88 = 0

⇒ x2 – 11x – 8x + 88 = 0

⇒ (x – 8) (x – 11) = 0

⇒ x = 8 or x = 11.

CRITERIA FOR SIMILARITY OF TWO TRIANGLES

Two triangles are said to be similar if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportional).

Thus, two triangles ABC and A'B'C' are similar if

(i) Basic Proportionality Theorem (Thales Theorem) and

(ii) Basic Proportionality Theorem (Thales Theorem).

In this section, we shall make use of the theorems discussed in earlier sections to derive some criteria for similar triangles which in turn will imply that either of the above two conditions can be used to define the similarity of two triangles.

Talk to Our counsellor