Q39
5 marksLong AnswerSection D

  1. (A) State and prove Basic Proportionality Theorem.

Triangles
Basic Proportionality Theorem - statement and proof
Official Answer

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


Given: △ABC with DE ∥ BC, intersecting AB at D and AC at E.

To Prove: AD/DB=AE/ECAD/DB = AE/EC.

Construction: Join BE and CD. Draw EM ⊥ AB and DN ⊥ AC.

Proof: Area(△ADE)=12×AD×EM\text{Area}(\triangle ADE) = \frac{1}{2} \times AD \times EM and Area(△BDE)=12×DB×EM\text{Area}(\triangle BDE) = \frac{1}{2} \times DB \times EM, so Area(△ADE)/Area(△BDE)=AD/DB\text{Area}(\triangle ADE)/\text{Area}(\triangle BDE) = AD/DB …(1). Similarly Area(△ADE)/Area(△CDE)=AE/EC\text{Area}(\triangle ADE)/\text{Area}(\triangle CDE) = AE/EC …(2). Since △BDE and △CDE lie on the same base DE between the same parallels DE and BC, Area(△BDE)=Area(△CDE)\text{Area}(\triangle BDE) = \text{Area}(\triangle CDE) …(3). From (1), (2), (3): AD/DB=AE/ECAD/DB = AE/EC. Hence proved.

Basic Proportionality TheoremThales theoremDE parallel to BCAD/DB = AE/ECsame base same parallelsequal areasperpendicular constructionratio of division

Marking Scheme

  • 11 mark: correct statement of the theorem (line parallel to one side divides the other two sides in the same ratio).
  • 21 mark: correct Given, To Prove and figure with the construction (joining BE, CD and drawing perpendiculars EM, DN).
  • 31 mark: obtaining Area(△ADE)/Area(△BDE)=AD/DB\text{Area}(\triangle ADE)/\text{Area}(\triangle BDE) = AD/DB using the common height EM.
  • 41 mark: obtaining Area(△ADE)/Area(△CDE)=AE/EC\text{Area}(\triangle ADE)/\text{Area}(\triangle CDE) = AE/EC using the common height DN.
  • 51 mark: stating Area(△BDE)=Area(△CDE)\text{Area}(\triangle BDE) = \text{Area}(\triangle CDE) (same base DE, between same parallels) and concluding AD/DB=AE/ECAD/DB = AE/EC.

Hint

Join BE and CD, drop perpendiculars EM⊥ABEM \perp AB and DN⊥ACDN \perp AC, compare areas of △ADE with △BDE and △CDE, then use that △BDE and △CDE have equal areas (same base DE, between parallels DE and BC).

Quick Oral Answer

The Basic Proportionality Theorem states that a line drawn parallel to one side of a triangle divides the other two sides in the same ratio; it is proved by joining BE and CD, comparing the areas of triangle ADE with triangles BDE and CDE using common heights, and using that BDE and CDE have equal areas as they share base DE between the parallels DE and BC.

Analysis & Explanation

The Basic Proportionality Theorem (Thales' Theorem) is the foundational area-method proof of the Triangles chapter.


Concept

  • CBSE expects four labelled parts: Given, To Prove, Construction, Proof.
  • Construction: join BE and CD; draw EM ⊥ AB and DN ⊥ AC to get common heights for area comparison.
  • Key step: △BDE and △CDE stand on the same base DE between the same parallels DE ∥ BC, so their areas are equal — this equality links the two area ratios and gives AD/DB=AE/ECAD/DB = AE/EC.

Common Mistakes

  • Omitting the construction (joining BE, CD) or the perpendiculars.
  • Skipping the justification for Area(△BDE) = Area(△CDE), even if the final ratio is stated correctly.

Real-World Relevance

  • Underlies scale drawings, map-making, similar-triangle range finders, and the shadow method for measuring heights of tall objects (the technique Thales reputedly used on the Great Pyramid).
  • Its converse (equal ratios on two sides ⇒ line parallel to the third side) is equally examinable.

Common Mistakes

  1. 1Omitting the construction (joining BE and CD and drawing the perpendiculars EM and DN) — without it the area ratios cannot be set up and marks are lost.
  2. 2Failing to justify why Area(△BDE)=Area(△CDE)\text{Area}(\triangle BDE) = \text{Area}(\triangle CDE); the reason 'same base DE and between the same parallels DE and BC' must be stated explicitly.
  3. 3Stating only the ratio result without labelling Given, To Prove, Construction and Proof separately, which does not meet CBSE's format expectation for a theorem proof.

Interesting Facts

The theorem is named after the ancient Greek mathematician Thales of Miletus (c. 624–546 BCE), who is said to have used similar triangles and shadow ratios to calculate the height of the Great Pyramid of Giza — one of the earliest recorded uses of indirect measurement.

The BPT and its converse together form the logical gateway to all three similarity criteria (AA, SSS, SAS) taught in Class 10; nearly every similarity proof in the chapter ultimately traces back to this single result.

The 'triangles on the same base and between the same parallels have equal areas' principle used in the proof is itself a corollary of the area formula ½ × base × height, since equal parallels guarantee equal perpendicular heights.

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Frequently Asked Questions

Why do we need to join BE and CD in the construction?

Joining BE and CD creates the triangles BDE and CDE. These auxiliary triangles are essential because they share base DE and lie between the parallels DE and BC, which makes their areas equal. Without this construction, there is no way to link the ratio AD/DBAD/DB to the ratio AE/ECAE/EC.

How is the Basic Proportionality Theorem different from its converse?

The BPT says: if a line is parallel to one side, it divides the other two sides proportionally (parallel ⇒ equal ratio). The converse says: if a line divides two sides of a triangle in the same ratio, then it is parallel to the third side (equal ratio ⇒ parallel). Both are examinable and are used for different types of problems.

Can the theorem be proved without using areas?

The NCERT-prescribed proof uses the area method, which is what CBSE expects for full marks. Alternative proofs using similar triangles exist, but since the similarity criteria themselves are derived from BPT, the area-based proof is the standard non-circular approach and should be used in the exam.