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BPT find-x problem: why the quadratic terms disappear

7 min read

Question

In triangle ABC, DE is parallel to BC. AD = x, DB = x − 2, AE = x + 2 and EC = x − 1. Find x.

BPT trick: how a vanishing quadratic makes you miss 2 solutions!Watch on YouTube

We need to form the correct Basic Proportionality Theorem ratio, solve the resulting equation and then test the value against the original segment lengths. No value will be accepted until both the algebra and the geometry have been checked.

ABCDEAD = xDB = x - 2AE = x + 2EC = x - 1DE parallel to BC
Because DE is parallel to BC, the two sides are cut proportionally. Match upper segment with upper segment and lower segment with lower segment.

Step 1: check which values of x are possible

In geometry, every side length must be strictly positive (l>0l > 0). Before solving, find the domain restriction for each of the four segments:

For segment AD>0  ⟹  x>0\text{For segment } AD > 0 \implies x > 0
For segment DB>0  ⟹  x−2>0  ⟹  x>2\text{For segment } DB > 0 \implies x - 2 > 0 \implies x > 2
For segment AE>0  ⟹  x+2>0  ⟹  x>−2\text{For segment } AE > 0 \implies x + 2 > 0 \implies x > -2
For segment EC>0  ⟹  x−1>0  ⟹  x>1\text{For segment } EC > 0 \implies x - 1 > 0 \implies x > 1
Combining conditions: x>0,  x>2,  x>−2,  x>1\text{Combining conditions: } x > 0,\; x > 2,\; x > -2,\; x > 1
Common restriction for all sides: x>2\text{Common restriction for all sides: } x > 2

The final value of xx must be strictly greater than 2 (x>2x > 2). This ensures all segment lengths remain positive and no denominator equals zero.

Step 2: form the BPT equation

Since line segment DE∥BCDE \parallel BC, we apply the Basic Proportionality Theorem (Thales' Theorem):

ADDB=AEEC[By Basic Proportionality Theorem, as DE∥BC]\frac{AD}{DB} = \frac{AE}{EC} \quad \text{[By Basic Proportionality Theorem, as } DE \parallel BC\text{]}
Given: AD=x,  DB=x−2,  AE=x+2,  EC=x−1\text{Given: } AD = x,\; DB = x - 2,\; AE = x + 2,\; EC = x - 1
Substitute values into ratio: xx−2=x+2x−1\text{Substitute values into ratio: } \frac{x}{x-2} = \frac{x+2}{x-1}

Step 3: cross-multiply without cancelling early

Cross-multiply the denominators and expand each side carefully:

xx−2=x+2x−1\frac{x}{x-2} = \frac{x+2}{x-1}
x(x−1)=(x−2)(x+2)[Cross-multiplication]x(x-1) = (x-2)(x+2) \quad \text{[Cross-multiplication]}
LHS expansion: x(x−1)=x⋅x−x⋅1\text{LHS expansion: } x(x-1) = x \cdot x - x \cdot 1
LHS: x(x−1)=x2−x\text{LHS: } x(x-1) = x^2 - x
RHS expansion: (x−2)(x+2)=x2+2x−2x−4\text{RHS expansion: } (x-2)(x+2) = x^2 + 2x - 2x - 4
RHS: (x−2)(x+2)=x2−4[Using identity (a−b)(a+b)=a2−b2]\text{RHS: } (x-2)(x+2) = x^2 - 4 \quad \text{[Using identity } (a-b)(a+b) = a^2 - b^2\text{]}
Equating LHS and RHS: x2−x=x2−4\text{Equating LHS and RHS: } x^2 - x = x^2 - 4
x2−x−x2=x2−4−x2[Subtract x2 from both sides]x^2 - x - x^2 = x^2 - 4 - x^2 \quad \text{[Subtract } x^2 \text{ from both sides]}
−x=−4-x = -4
−x−1=−4−1[Divide both sides by −1]\frac{-x}{-1} = \frac{-4}{-1} \quad \text{[Divide both sides by } -1\text{]}
x=4x = 4

Notice how the quadratic terms x2x^2 cancel out cleanly on both sides when x2x^2 is subtracted. This leaves a linear equation with one unique root.

Step 4: verify the candidate in the original data

Substitute x=4x = 4 back into all four segment expressions and check the ratios:

Segment AD=x=4 cm\text{Segment } AD = x = 4\text{ cm}
Segment DB=x−2=4−2=2 cm\text{Segment } DB = x - 2 = 4 - 2 = 2\text{ cm}
Segment AE=x+2=4+2=6 cm\text{Segment } AE = x + 2 = 4 + 2 = 6\text{ cm}
Segment EC=x−1=4−1=3 cm\text{Segment } EC = x - 1 = 4 - 1 = 3\text{ cm}
Condition check: x=4>2(All lengths are positive ✓)\text{Condition check: } x = 4 > 2 \quad (\text{All lengths are positive } \checkmark)
Left ratio: ADDB=42=2\text{Left ratio: } \frac{AD}{DB} = \frac{4}{2} = 2
Right ratio: AEEC=63=2\text{Right ratio: } \frac{AE}{EC} = \frac{6}{3} = 2
ADDB=AEEC=2(BPT ratio holds true ✓)\frac{AD}{DB} = \frac{AE}{EC} = 2 \quad (\text{BPT ratio holds true } \checkmark)

All four segment lengths are positive, no denominators are zero, and both ratios equal 2. The solution is completely verified.

Three checks that prevent BPT errors

  • Keep the ratio order consistent. Upper/lower must equal upper/lower, not upper/lower equals lower/upper.
  • Use brackets before expanding. In particular, preserve both factors in (x−2)(x+2)(x-2)(x+2).
  • Test segment positivity. An algebraic root is unusable if it makes a geometric length zero or negative.

Final step: state the verified answer

x=4\boxed{x = 4}

Young Math Brains trains you to match corresponding segments before it allows cross-multiplication, then checks every derived length.

Practise BPT questions interactively

Your turn

At x = 4, what are the two equal ratios AD:DB and AE:EC?

Tell us in the comments on YouTube or Instagram, or email connect@youngmathbrains.com. We read every one.

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