Question 222 marksVery Short AnswerSection B

  1. (A) In △ABC\triangle ABC, DE∥BCDE \parallel BC. If AD=xAD = x, DB=x−2DB = x - 2, AE=x+2AE = x + 2 and EC=x−1EC = x - 1, then find the value of x.

Triangles
Basic Proportionality Theorem

Official answer

x=4x = 4. By BPT, ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}, so xx−2=x+2x−1\frac{x}{x - 2} = \frac{x + 2}{x - 1}; cross-multiplying gives x2−x=x2−4x^2 - x = x^2 - 4, which simplifies to x=4x = 4.

Answer keywords:Basic Proportionality TheoremThales theoremDE parallel BCAD/DB = AE/ECcross multiplyx = 4difference of squares

Marking scheme

  1. 1.½ mark: Stating the Basic Proportionality Theorem / writing ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}.
  2. 2.½ mark: Correct substitution xx−2=x+2x−1\frac{x}{x-2} = \frac{x+2}{x-1}.
  3. 3.½ mark: Correct cross-multiplication and expansion x2−x=x2−4x^2 - x = x^2 - 4.
  4. 4.½ mark: Correct final value x=4x = 4 (a verification step is appreciated but not mandatory).

Hint

DE∥BCDE \parallel BC means ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}. Cross-multiply — the x2x^2 terms will cancel, leaving a simple linear equation.

Quick oral answer

Because DE is parallel to BC, the Basic Proportionality Theorem gives ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}; substituting the lengths and cross-multiplying makes the x2x^2 terms cancel, leaving x=4x = 4, which checks out since both ratios equal 2.

Common mistakes

  1. 1.Pairing the wrong segments — writing ADAE=DBEC\frac{AD}{AE} = \frac{DB}{EC} or ADDB=ECAE\frac{AD}{DB} = \frac{EC}{AE} instead of the correct ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}.
  2. 2.Forgetting that (x+2)(x−2)=x2−4(x + 2)(x - 2) = x^2 - 4 (difference of squares) and instead expanding it incorrectly, which prevents the x2x^2 terms from cancelling.
  3. 3.Stopping at the quadratic stage and hunting for two roots, when the equation actually reduces to a single linear equation with x=4x = 4.

Analysis and explanation

A textbook application of the Basic Proportionality Theorem (Thales' Theorem) with an algebraic twist.


Concept

  • DE∥BC in △ABC⇒ADDB=AEECDE \parallel BC \text{ in } \triangle ABC \Rightarrow \frac{AD}{DB} = \frac{AE}{EC} (ratio of segments, not full sides).
  • Substituting the given expressions and cross-multiplying causes the quadratic terms to cancel, collapsing the equation to a simple linear one.

Key points

  • x(x−1)=(x+2)(x−2)⇒x2−x=x2−4⇒x=4x(x - 1) = (x + 2)(x - 2) \Rightarrow x^2 - x = x^2 - 4 \Rightarrow x = 4.
  • Checking: AD=4,DB=2,AE=6,EC=3AD = 4, DB = 2, AE = 6, EC = 3; both ratios equal 2, confirming DE∥BCDE \parallel BC.

Common mistakes

  • Writing ADAB=AEAC\frac{AD}{AB} = \frac{AE}{AC} (full sides) instead of the correct segment ratio ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC}.
  • If the terms don't cancel to a clean linear equation, it signals the segments were paired incorrectly.

Interesting facts

The Basic Proportionality Theorem is named after Thales of Miletus (c. 624–546 BCE), who reportedly used similar-triangle proportions to measure the height of the Great Pyramid from its shadow.

BPT's converse is equally examined: if a line divides two sides of a triangle in the same ratio, it must be parallel to the third side — the check we performed (both ratios = 2) is essentially this converse.

Question details

Q22 of 53 questions

Board
CBSE
Class
Class 10
Subject
Mathematics (Basic)
Year
2026
Section
Section B
Marks
2
Type
Very Short Answer
Chapter
Triangles
Topic
Basic Proportionality Theorem
Bloom's level
apply
Syllabus topic
Unit 4: Geometry — Triangles: Basic Proportionality Theorem (Thales' Theorem)
Author
CBSE
Reviewed by
Padhantu Academic Team

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Frequently asked questions

Which ratio does the Basic Proportionality Theorem give here?

Since DE∥BCDE \parallel BC cuts sides AB and AC, BPT gives ADDB=AEEC\frac{AD}{DB} = \frac{AE}{EC} (the parts of each side, not the whole sides). This becomes xx−2=x+2x−1\frac{x}{x-2} = \frac{x+2}{x-1}.

Why does the quadratic reduce to a linear equation?

Cross-multiplying gives x(x−1)=(x+2)(x−2)x(x-1) = (x+2)(x-2). The right side is a difference of squares = x2−4x^2 - 4, and the left is x2−xx^2 - x. The x² terms cancel, leaving −x=−4-x = -4, so x=4x = 4.

How can I check the answer x=4x = 4 is correct?

Substitute back: AD=4,DB=2,AE=6,EC=3AD = 4, DB = 2, AE = 6, EC = 3. Then ADDB=2\frac{AD}{DB} = 2 and AEEC=2\frac{AE}{EC} = 2. Equal ratios confirm DE∥BCDE \parallel BC, so x=4x = 4 is verified.