Q22

2 marksVery Short AnswerSection B

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

Q: 22. (A) In $\triangle ABC$, $DE \parallel BC$. If $AD = x$, $DB = x - 2$, $AE = x + 2$ and $EC = x - 1$, then find the value of x.

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

Q: Why does the quadratic reduce to a linear equation?

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

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

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

Triangles

Basic Proportionality Theorem

Official Answer

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

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

Marking Scheme

1½ mark: Stating the Basic Proportionality Theorem / writing $\frac{AD}{DB} = \frac{AE}{EC}$ .
2½ mark: Correct substitution $\frac{x}{x-2} = \frac{x+2}{x-1}$ .
3½ mark: Correct cross-multiplication and expansion $x^2 - x = x^2 - 4$ .
4½ mark: Correct final value $x = 4$ (a verification step is appreciated but not mandatory).

Hint

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

Quick Oral Answer

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

Analysis & Explanation

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

Concept

$DE \parallel BC \text{ in } \triangle ABC \Rightarrow \frac{AD}{DB} = \frac{AE}{EC}$ (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

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

Common mistakes

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

Common Mistakes

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

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.

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Frequently Asked Questions

Which ratio does the Basic Proportionality Theorem give here?

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

Why does the quadratic reduce to a linear equation?

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

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

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