Basic Proportionality Theorem
A key idea in geometry is the Basic Proportionality Theorem, or BPT Theorem. It was initially presented by the Greek mathematician Thales and is frequently employed in the study of triangles and related shapes. The theorem clarifies how a line drawn parallel to one triangle side proportionally divides the other two sides.
Table of Contents
- What is the Basic Proportionality Theorem?
- Proof of the Basic Proportionality Theorem
- Converse of Basic Proportionality Theorem
- Proof of the Converse Theorem
- Solved Examples of the Basic Proportionality Theorem
- Practice Questions
- Conclusion
- Frequently Asked Questions (FAQs)
What is the Basic Proportionality Theorem?
The BPT theorem, sometimes referred to as the Thales Theorem after its discoverer, is a geometrical result that connects parallel lines and proportional triangle segments.
The Basic Proportionality Theorem statement:
It is stated as:
If a line is drawn parallel to one side of a triangle and intersects the other two sides, it divides those two sides in the same ratio.
Mathematical Form:
In triangle ABC, if DE is drawn parallel to BC, intersecting AB at D and AC at E, then:
AD/DB = AE/EC
A fundamental aspect of geometry is the property that the ratios of corresponding sides are equal. The basic proportionality theorem, or BPT theorem, states that proportional division is guaranteed by DE ∥ BC.
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AD: DB:: AE: EC.
This theorem forms the basis for exploring the basic proportionality theorem converse later, and for solving geometric problems using proportionality.
Proof of the Basic Proportionality Theorem
Given:
In triangle ABC, a line DE is drawn parallel to side BC. The line intersects AB at point D and AC at point E.
Construction:
Draw triangle ABC. Draw a line DE ∥ BC, intersecting AB at D and AC at E.
From point D, draw DM ⊥ AC.
From point E, draw EN ⊥ AB.
Let M and N be the feet of perpendiculars from D and E, respectively.
To Prove:
AD / DB = AE / EC
Proof:
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Consider triangles ADE and DBE.
Both have a common height from point E (EN) and lie between the same parallels. -
Area of triangle ADE = (1/2) × AD × EN
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Area of triangle DBE = (1/2) × DB × EN
-
Therefore,
Area(ΔADE) / Area(ΔDBE) = AD / DB ...(1) -
Now consider triangles ADE and EDC.
Both have a common height from point D (DM) and lie between the same parallels. -
Area of triangle ADE = (1/2) × AE × DM
-
Area of triangle EDC = (1/2) × EC × DM
-
Therefore,
Area(ΔADE) / Area(ΔEDC) = AE / EC ...(2) -
From equations (1) and (2), we equate the area ratios:
AD / DB = AE / EC
Proved:
∴ AD / DB = AE / EC
Conclusion:
Hence, it is proved that if a line is drawn parallel to one side of a triangle and intersects the other two sides, it divides those sides in the same ratio.
This completes the proof of the Basic Proportionality Theorem.
Converse of Basic Proportionality Theorem
Statement of the basic proportionality theorem converse:
If a line divides any two sides of a triangle in the same ratio, then it is parallel to the third side.
In triangle ABC, if a line intersects AB at D and AC at E such that:
AD / DB = AE / EC
Then line DE ∥ BC.
Proof of the Converse Theorem
Given:
In triangle ABC, points D and E lie on sides AB and AC, respectively, such that:
AD / DB = AE / EC
Construction:
Draw triangle ABC. Mark points D on AB and E on AC such that AD / DB = AE / EC.
Now, draw a line D'E' through point D such that D'E' ∥ BC, and let it intersect AC at E′.
To Prove:
Line DE ∥ BC
Proof:
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In triangle ABC, since D′E′ ∥ BC (by construction), by the Basic Proportionality Theorem:
AD / DB = AE′ / E′C ...(1) -
But it is given that:
AD / DB = AE / EC ...(2) -
From (1) and (2), we have:
AE / EC = AE′ / E′C
This means point E and point E′ must coincide.
Therefore, line DE and line D′E′ are the same line.
Hence, DE ∥ BC.
Proved:
∴ DE ∥ BC
Hence, it is proved that if a line divides any two sides of a triangle in the same ratio, then it is parallel to the third side.
This completes the proof of the Converse of the Basic Proportionality Theorem.
Solved Examples of the Basic Proportionality Theorem
Example 1:
In triangle ABC, a line DE is drawn parallel to side BC.
It intersects AB at D and AC at E.
If AD = 3 cm, DB = 6 cm, and AE = 2 cm, find the length of EC.
Solution:
Since DE ∥ BC, by the Basic Proportionality Theorem:
AD / DB = AE / EC
Substitute the known values:
3 / 6 = 2 / EC
⇒ 1 / 2 = 2 / EC
Cross-multiplying:
1 × EC = 2 × 2
⇒ EC = 4 cm
Example 2:
In triangle PQR, a line ST is drawn parallel to QR.
It intersects PQ at S and PR at T.
If PS = 4 cm, SQ = 6 cm, and PT = 5 cm, find the length of TR.
Solution:
Since ST ∥ QR, by the Basic Proportionality Theorem:
PS / SQ = PT / TR
Substitute the values:
4 / 6 = 5 / TR
⇒ 2 / 3 = 5 / TR
Cross-multiplying:
2 × TR = 15
⇒ TR = 15 / 2 = 7.5 cm
Example 3:
In triangle XYZ, a line AB is drawn parallel to YZ.
It intersects XY at A and XZ at B.
If XA = 6 cm, AY = 9 cm, and XB = 4 cm, find the length of BZ.
Solution:
Since AB ∥ YZ, by the Basic Proportionality Theorem:
XA / AY = XB / BZ
Substitute the values:
6 / 9 = 4 / BZ
⇒ 2 / 3 = 4 / BZ
Cross-multiplying:
2 × BZ = 12
⇒ BZ = 6 cm
Example 4:
In triangle LMN, a line PQ is drawn parallel to MN.
It intersects LM at P and LN at Q.
If LP = 5 cm, PM = 10 cm, and LQ = 4 cm, find the value of QN.
Solution:
Since PQ ∥ MN, by the Basic Proportionality Theorem:
LP / PM = LQ / QN
Substitute the values:
5 / 10 = 4 / QN
⇒ 1 / 2 = 4 / QN
Cross-multiplying:
1 × QN = 2 × 4
⇒ QN = 8 cm
Example 5:
In triangle ABC, D and E are points on AB and AC such that DE ∥ BC.
If AD = 2 cm, AE = 3 cm, and DB = 6 cm, find the length of EC.
Solution:
Since DE ∥ BC,
AD / DB = AE / EC
Substitute values:
2 / 6 = 3 / EC
⇒ 1 / 3 = 3 / EC
Cross-multiplying:
1 × EC = 3 × 3
⇒ EC = 9 cm
Practice Questions
Try solving the following questions to test your understanding of the basic proportionality theorem:
Q1: In triangle PQR, ST || QR, PS = 4 cm, SQ = 8 cm, PT = 6 cm. Find TR.
Q2: In triangle ABC, D and E are on AB and AC such that DE || BC.
If AD = 5, DB = 10, AE = x, and EC = 6, find the value of x.
Q3: In triangle DEF, line XY divides DE and DF in the same ratio. What does the Basic Proportionality Theorem Converse tell you?
Q4: DE is drawn parallel to BC in triangle ABC. If AD = 2, DB = 3, AE = 4, find EC.
Use the proof of the basic proportionality theorem logic to solve these.
Conclusion
In triangle geometry, the Basic Proportionality Theorem (BPT) is a fundamental idea. It illustrates how a line drawn parallel to one triangle side divides the other two sides in the same proportion. This theorem
was discovered by Thales and is essential to the solution of challenging geometric problems in addition to improving our comprehension of proportional relationships.
The theorem and its converse provide us with strong tools for geometric reasoning, constructions, and proofs.
You now possess the skills necessary to confidently apply the Basic Proportionality Theorem in a variety of situations thanks to the solved examples and practice questions. Continue honing your problem-solving speed and conceptual clarity.
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Frequently Asked Questions (FAQs)
1. Who discovered the Basic Proportionality Theorem?
The Greek mathematician Thales made the discovery of the Basic Proportionality Theorem. As a result, Thales' Theorem is another name for it.
2. What is the Basic Proportionality Theorem?
It states that if a line is drawn parallel to one side of a triangle and intersects the other two sides, it divides those sides in the same ratio.
3. What is the mathematical expression of the Basic Proportionality Theorem?
In triangle ABC, if DE ∥ BC and intersects AB at D and AC at E, then:
AD / DB = AE / EC
4. What is the Converse of the Basic Proportionality Theorem?
If a line divides two sides of a triangle in the same ratio, then it is parallel to the third side.
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