MathsCBSE Class 10 · Triangles

Basic Proportionality Theorem (Thales Theorem) Class 10

Q: Do I need to learn the proof of BPT?

Yes. It is the only theorem in the Triangles chapter that the syllabus asks you to prove. Practise writing it with given, to prove, construction and reasons.

Q: Do I need to prove the converse of BPT?

No. The converse is stated without proof. You only quote and use it.

Q: Can I use BPT to find the length of DE?

No. BPT only links the pieces of the two cut sides. For DE, prove the small and big triangles similar by AA and use the whole side ratio.

Q: Does the BPT proof need any special area result?

No. The area step in the proof uses only the formula half × base × height, and the fact that triangles on the same base and between the same parallels have equal areas.

Updated 14 October 2026 · Fact-checked

The Basic Proportionality Theorem says that a line drawn parallel to one side of a triangle, cutting the other two sides at distinct points, divides those sides in the same ratio. The board asks you to prove it as a long answer and to use it to find unknown lengths or values of x.

Understand Basic Proportionality Theorem (Thales Theorem)

Start with a triangle ABC. Draw a line parallel to the base BC so that it cuts AB at D and AC at E. The Basic Proportionality Theorem (BPT), also called Thales theorem, says the two sides are cut in the same ratio: $\frac{A D}{D B} = \frac{A E}{E C}$ . If D is one-third of the way down AB, then E is one-third of the way down AC.

Why it works. The proof uses areas, not similarity. Join BE and CD. Triangles ADE and BDE have bases AD and DB on the same line AB and the same top vertex E, so they share one height. Their areas are therefore in the ratio $A D : D B$ . In the same way, triangles ADE and DEC share a height from D to AC, so their areas are in the ratio $A E : E C$ . The key step is that triangles BDE and DEC stand on the same base DE and lie between the same parallels DE and BC, so they have equal areas. Two ratios with the same numerator and equal denominators must be equal.

Conditions. The line must be parallel to a side, and it must meet the other two sides at distinct points (not at a vertex). Without the parallel line, the ratios need not be equal. BPT is the only theorem in this chapter you must prove in the exam.

Useful forms. From $\frac{A D}{D B} = \frac{A E}{E C}$ you can flip both sides, add 1 and get $\frac{A B}{A D} = \frac{A C}{A E}$ , so $\frac{A D}{A B} = \frac{A E}{A C}$ and $\frac{D B}{A B} = \frac{E C}{A C}$ . Pick the form that uses the lengths given in the question.

Connections. BPT only relates pieces of the two cut sides. To find the parallel segment DE itself, you need $△ A D E \sim △ A B C$ (AA), which gives $\frac{D E}{B C} = \frac{A D}{A B}$ . Its converse (ratio implies parallel) and the similarity criteria have their own pages. BPT also explains why a line parallel to the parallel sides of a trapezium cuts the other two sides proportionally.

Basic Proportionality Theorem (Thales Theorem): formulas to remember

Basic Proportionality Theorem: $D E ∥ B C \Rightarrow \frac{A D}{D B} = \frac{A E}{E C}$; D on AB, E on AC. Part to part on both sides.
Part-to-whole form: $\frac{A D}{A B} = \frac{A E}{A C}$; Use when a whole side is given.
Lower part to whole: $\frac{D B}{A B} = \frac{E C}{A C}$; Also follows from BPT.
Area step in the proof: $\frac{a r ( A D E )}{a r ( B D E )} = \frac{A D}{D B}, \frac{a r ( A D E )}{a r ( D E C )} = \frac{A E}{E C}, a r (B D E) = a r (D E C)$; Equal areas because both are on base DE between parallels DE and BC.
Parallel segment (via similarity, not BPT): $\frac{D E}{B C} = \frac{A D}{A B}$; Needs $△ A D E \sim △ A B C$ by AA; always use the whole side AB.

How to solve Basic Proportionality Theorem (Thales Theorem) questions

Every BPT question follows the same pattern: spot the parallel line, match the pieces, solve.

1Identify the triangle and the line parallel to one of its sides; name where it cuts the other two sides.
2Write the reason: "In $△ A B C$ , $D E ∥ B C$ , so by BPT".
3Choose the form: part to part if pieces are given, part to whole if a whole side is given.
4Keep vertex order: the piece next to the common vertex goes on top on both sides.
5Cross-multiply and solve. Expand carefully: the equation may reduce to a linear equation or a quadratic. Reject any root that makes a length zero or negative.
6Check every length is positive and substitute back into the ratio.
7If the parallel segment itself is asked, switch to similarity and use $\frac{D E}{B C} = \frac{A D}{A B}$ .

Basic Proportionality Theorem (Thales Theorem) in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Basic Proportionality Theorem

Suppose a triangular sloping roof gable ABC is built for a school shed, with A at the top and BC the horizontal beam, 5.4 m long. Two horizontal shelf beams DE and FG are fixed parallel to BC, with D and F on the slanting edge AB and E and G on edge AC, in the order A, D, F, B and A, E, G, C. Along AB, $A D = 1.5$ m, $D F = 1.8$ m and $F B = 1.2$ m. The whole edge AC is 6 m long.

Find where each beam meets AC (the lengths AE, EG and GC), and find the lengths of both beams.

Try it first, then show the solutionSolution

AB $= 1.5 + 1.8 + 1.2 = 4.5$ m, and $A F = 1.5 + 1.8 = 3.3$ m.
$D E ∥ B C$ , so by BPT $\frac{A D}{A B} = \frac{A E}{A C}$ : $\frac{1.5}{4.5} = \frac{A E}{6}$ , so $A E = 2$ m.
$F G ∥ B C$ , so $\frac{A F}{A B} = \frac{A G}{A C}$ : $\frac{3.3}{4.5} = \frac{A G}{6}$ , so $A G = 4.4$ m.
So $E G = 4.4 - 2 = 2.4$ m and $GC = 6 - 4.4 = 1.6$ m. Check: the pieces of AC (2, 2.4, 1.6) are $\frac{4}{3}$ times the pieces of AB (1.5, 1.8, 1.2).
$△ A D E \sim △ A B C$ (AA: $∠ A$ common, corresponding angles), so $D E = \frac{1.5}{4.5} \times 5.4 = 1.8$ m.
Similarly $△ A F G \sim △ A B C$ , so $F G = \frac{3.3}{4.5} \times 5.4 = 3.96$ m.

Answer: AE = 2 m, EG = 2.4 m, GC = 1.6 m; the upper beam DE is 1.8 m and the lower beam FG is 3.96 m long.

Memory hook: A level beam in a sloping roof cuts both slopes in the same ratio.

Tricks to remember Basic Proportionality Theorem (Thales Theorem)

Short memory aids. Say each one aloud once, then use it on a question the same day.

Top over bottom, both sides
Read each side from the vertex A down: top piece over bottom piece on AB equals top piece over bottom piece on AC.
Use it when: Stops you from flipping one ratio.
Parallel gives proportion
BPT: Parallel to Proportion. Converse: Proportion to Parallel.
Use it when: Decide which result to quote.
Proof in three words
Join, Height, Same base. Join BE and CD; use common heights; BDE and DEC on same base between same parallels.
Use it when: Recall the proof structure in the exam.
DE needs the whole
For the parallel segment, always divide by the whole side: $\frac{D E}{B C} = \frac{A D}{A B}$ , never $\frac{A D}{D B}$ .
Use it when: Avoid the most common length error.

Basic Proportionality Theorem (Thales Theorem): solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Assertion-reason: which ratio does BPT give? · 1 mark · Assertion-reasonBoard level

Assertion (A): In $△ A B C$ , $D E ∥ B C$ with D on AB and E on AC. If $A D = 2$ cm, $D B = 3$ cm and $B C = 10$ cm, then $D E = 4$ cm.
Reason (R): If $D E ∥ B C$ , then $\frac{D E}{B C} = \frac{A D}{D B}$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

$△ A D E \sim △ A B C$ (AA), so $\frac{D E}{B C} = \frac{A D}{A B} = \frac{2}{5}$ and $D E = 4$ cm. A is true.
R uses DB instead of the whole side AB: it would give $D E = \frac{2}{3} \times 10 \approx 6.67$ cm. R is false. (1)

Answer: (c) A is true but R is false.

Example 2: Solve for x · 3 marks · Short answerBoard level

In $△ A B C$ , $D E ∥ B C$ with D on AB and E on AC. $A D = (4 x - 3)$ cm, $D B = (3 x - 1)$ cm, $A E = (8 x - 7)$ cm and $E C = (5 x - 3)$ cm. Find $x$ and AB.

Try it first, then show the solutionSolution

By BPT, $\frac{A D}{D B} = \frac{A E}{E C}$ . (½)
$(4 x - 3) (5 x - 3) = (3 x - 1) (8 x - 7)$ . (½)
$20 x^{2} - 27 x + 9 = 24 x^{2} - 29 x + 7$ . (½)
$4 x^{2} - 2 x - 2 = 0$ , so $2 x^{2} - x - 1 = 0$ , $(2 x + 1) (x - 1) = 0$ . (½)
$x = 1$ or $x = - \frac{1}{2}$ ; $x = - \frac{1}{2}$ gives $A D < 0$ , rejected. (½)
$x = 1$ : AD = 1, DB = 2, AE = 1, EC = 2; ratio $\frac{1}{2}$ both. AB = 3 cm. (½)

Answer: x = 1, AB = 3 cm

Example 3: BPT twice in one figure · 3 marks · Short answerHarder board level

In $△ A B C$ , D is on AB and E is on AC with $D E ∥ B C$ . F is on AD with $F E ∥ D C$ (here DC is the segment joining D to C). If $A F = 4$ cm and $F D = 2$ cm, find $\frac{A D}{D B}$ and DB.

Try it first, then show the solutionSolution

In $△ A D C$ , $F E ∥ D C$ , so by BPT $\frac{A F}{F D} = \frac{A E}{E C}$ . (1)
In $△ A B C$ , $D E ∥ B C$ , so by BPT $\frac{A D}{D B} = \frac{A E}{E C}$ . (1)
Hence $\frac{A D}{D B} = \frac{A F}{F D} = \frac{4}{2} = 2$ . (½)
AD = 6 cm, so DB = 3 cm. (½)

Answer: AD : DB = 2 : 1, DB = 3 cm

Example 4: Prove BPT and apply it · 5 marks · Long answerBoard level

(i) Prove that if a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio.
(ii) In $△ X Y Z$ , $M N ∥ Y Z$ with M on XY and N on XZ. $X M = 5$ cm, XY = 12.5 cm and $N Z = 6$ cm. Find XN.

Try it first, then show the solutionSolution

Given: $△ A B C$ , $D E ∥ B C$ , D on AB, E on AC. To prove: $\frac{A D}{D B} = \frac{A E}{E C}$ . Construction: join BE, CD; draw $E N ⊥ A B$ , $D M ⊥ A C$ . (½)
$a r (A D E) = \frac{1}{2} A D \cdot E N$ , $a r (B D E) = \frac{1}{2} D B \cdot E N$ , so $\frac{a r ( A D E )}{a r ( B D E )} = \frac{A D}{D B}$ . (1)
$a r (A D E) = \frac{1}{2} A E \cdot D M$ , $a r (D E C) = \frac{1}{2} E C \cdot D M$ , so $\frac{a r ( A D E )}{a r ( D E C )} = \frac{A E}{E C}$ . (½)
$△ B D E$ and $△ D E C$ are on the same base DE and between the same parallels DE and BC, so $a r (B D E) = a r (D E C)$ . (1)
Therefore $\frac{A D}{D B} = \frac{A E}{E C}$ . Proved. (½)
(ii) $M Y = 12.5 - 5 = 7.5$ cm. By BPT $\frac{X M}{M Y} = \frac{X N}{N Z}$ . (½)
$\frac{5}{7.5} = \frac{X N}{6}$ , so $X N = \frac{2}{3} \times 6 = 4$ cm. (1)

Answer: (i) Proved. (ii) XN = 4 cm

Example 5: Diagonals of a trapezium · 5 marks · Long answerBoard level

(i) ABCD is a trapezium with $A B ∥ D C$ . Its diagonals AC and BD meet at O. Prove that $\frac{A O}{O C} = \frac{B O}{O D}$ .
(ii) In such a trapezium, $A O = (x + 3)$ cm, $O C = 3 x$ cm, $B O = (x + 1)$ cm and $O D = 2 x$ cm. Find $x$ .

Try it first, then show the solutionSolution

Construction: through O draw a line parallel to AB meeting AD at E. Since $A B ∥ D C$ , also $E O ∥ D C$ . (1)
In $△ A D C$ , $E O ∥ D C$ , so by BPT $\frac{A E}{E D} = \frac{A O}{O C}$ . (1)
In $△ D A B$ , $E O ∥ A B$ , so by BPT $\frac{D E}{E A} = \frac{D O}{O B}$ , that is $\frac{A E}{E D} = \frac{B O}{O D}$ . (½)
Hence $\frac{A O}{O C} = \frac{B O}{O D}$ . Proved. (½)
(ii) $\frac{x + 3}{3 x} = \frac{x + 1}{2 x}$ , so $2 x (x + 3) = 3 x (x + 1)$ . (½)
$2 x^{2} + 6 x = 3 x^{2} + 3 x$ , so $x^{2} - 3 x = 0$ , $x (x - 3) = 0$ . (½)
$x = 0$ gives OC = 0, which is impossible, so $x = 3$ . (½)
Check: AO = 6, OC = 9, BO = 4, OD = 6; $\frac{6}{9} = \frac{4}{6} = \frac{2}{3}$ . (½)

Answer: (i) Proved. (ii) x = 3

Example 6: Case study: a ladder rack · 4 marks · Case-basedHOTS / competency

Suppose a carpenter makes a triangular stand ABC for a school science display, with A at the top and BC on the floor, 80 cm long. Two horizontal shelves PQ and RS are fixed parallel to BC, with P and R on AB, Q and S on AC, in the order A, P, R, B and A, Q, S, C. On AB: $A P = 30$ cm, $P R = 45$ cm, $R B = 25$ cm. On AC, the carpenter only noted that $S C = x$ cm and $A C = (x + 90)$ cm.
(i) Find x and AC. (1)
(ii) Find AQ and QS. (1)
(iii) Find the ratio QS : SC and the lengths of both shelves. (2)

Try it first, then show the solutionSolution

(i) AB = 100 cm and $A R = 75$ cm. $R S ∥ B C$ , so by BPT $\frac{R B}{A B} = \frac{S C}{A C}$ : $\frac{25}{100} = \frac{x}{x + 90}$ , so $4 x = x + 90$ , $x = 30$ and $A C = 120$ cm. (1)
(ii) $P Q ∥ B C$ : $\frac{A P}{A B} = \frac{A Q}{A C}$ , so $A Q = \frac{30}{100} \times 120 = 36$ cm. $A S = 120 - 30 = 90$ cm, so $QS = 54$ cm. (1)
(iii) $QS : S C = 54 : 30 = 9 : 5$ (the same as $P R : R B = 45 : 25$ ). (1)
$△ A P Q \sim △ A B C$ and $△ A R S \sim △ A B C$ (AA), so $P Q = 0.3 \times 80 = 24$ cm and $R S = 0.75 \times 80 = 60$ cm. (1)

Answer: (i) x = 30, AC = 120 cm (ii) AQ = 36 cm, QS = 54 cm (iii) QS : SC = 9 : 5; shelves PQ = 24 cm and RS = 60 cm.

Example 7: Trapezium: BPT, AA and a line through the diagonals · 5 marks · Long answerHarder board level

ABCD is a trapezium with $A B ∥ D C$ , $A B = 15$ cm and $D C = 10$ cm. Its diagonals meet at O. A line through O parallel to AB meets AD at E and BC at F. Find $A E : E D$ , EO, OF and EF.

Try it first, then show the solutionSolution

$△ A O B \sim △ C O D$ (AA: alternate angles $∠ O A B = ∠ O C D$ , vertically opposite angles at O), so $\frac{A O}{O C} = \frac{A B}{D C} = \frac{3}{2}$ . (1)
EO $∥$ DC, so by BPT in $△ A D C$ : $\frac{A E}{E D} = \frac{A O}{O C} = \frac{3}{2}$ . (1)
$△ A E O \sim △ A D C$ (AA), so $\frac{E O}{D C} = \frac{A O}{A C} = \frac{3}{5}$ and $E O = 6$ cm. (1)
$△ C O F \sim △ C A B$ (AA, OF $∥$ AB), so $\frac{O F}{A B} = \frac{C O}{C A} = \frac{2}{5}$ and $O F = 6$ cm. (1)
EF = 12 cm, and O is its midpoint. (1)

Answer: AE : ED = 3 : 2, EO = OF = 6 cm and EF = 12 cm.

Where students lose marks in Basic Proportionality Theorem (Thales Theorem)

The slips examiners see most often, and what to write instead.

Mistake: Using BPT to find the parallel segment
BPT does not mention DE at all.
Wrong working
$\frac{D E}{B C} = \frac{A D}{D B}$
Right working
$△ A D E \sim △ A B C$ (AA), so $\frac{D E}{B C} = \frac{A D}{A B}$
Do this: For DE, prove similarity and use the whole side.
Usually costs: Usually the whole answer mark.
Mistake: Mixing part and whole
Comparing a piece on one side with a whole on the other.
Wrong working
$\frac{A D}{D B} = \frac{A E}{A C}$
Right working
$\frac{A D}{D B} = \frac{A E}{E C}$ or $\frac{A D}{A B} = \frac{A E}{A C}$
Do this: Write both ratios in the same form.
Usually costs: 1 to 2 marks.
Mistake: Not rejecting an invalid root
A root of the equation can make a length negative.
Wrong working
x = 1 or x = −½, both accepted
Right working
x = −½ makes AD negative, so x = 1 only
Do this: Substitute each root into every length.
Usually costs: ½ to 1 mark.
Mistake: Missing reason for equal areas in the proof
This is the key step examiners look for.
Wrong working
ar(BDE) = ar(DEC)
Right working
ar(BDE) = ar(DEC), same base DE and between the same parallels DE and BC
Do this: Always write the full reason.
Usually costs: 1 mark in the proof.

Exam tips

In a proof, write Given, To prove, Construction and the figure description before the steps.
Always quote the reason "by BPT" with the parallel lines named; a bare ratio loses the reason mark.
When expressions in x are given, expand carefully. The equation may reduce to a linear equation or a quadratic; reject any root that makes a length zero or negative.
If a question asks for the parallel segment, expect to use AA similarity, not BPT.
Check answers by substituting back into both ratios.

Practice questions on Basic Proportionality Theorem (Thales Theorem)

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 3 marks · Short answerHarder board level
In $△ A B C$ , $D E ∥ B C$ with D on AB and E on AC. $A D = x$ cm, $D B = (2 x - 3)$ cm, $A E = (x + 1)$ cm and $E C = (3 x - 5)$ cm. (i) Find x, rejecting any impossible value. (ii) F is a point on DB with $D F = 1$ cm, and $F G ∥ B C$ meets AC at G. Find EG.
Hint
After finding x, use BPT again with AF and the whole side AB.
Answer
(i) $x (3 x - 5) = (2 x - 3) (x + 1)$ gives $3 x^{2} - 5 x = 2 x^{2} - x - 3$ , so $x^{2} - 4 x + 3 = 0$ and $x = 1$ or $3$ . $x = 1$ makes $D B = - 1$ , so reject it: $x = 3$ , giving AD = 3, DB = 3, AE = 4, EC = 4, so AB = 6 cm and AC = 8 cm. (ii) $A F = 4$ cm. By BPT, $\frac{A F}{A B} = \frac{A G}{A C}$ : $A G = \frac{4}{6} \times 8 = \frac{16}{3}$ cm, so $E G = \frac{16}{3} - 4 = \frac{4}{3}$ cm.
Q2 · 3 marks · Short answerHarder board level
ABCD is a trapezium with $A B ∥ D C$ . E is on AD and F is on BC with $E F ∥ A B$ . Diagonal AC cuts EF at G. Prove that $\frac{A E}{E D} = \frac{B F}{F C}$ . If $A E = 6$ cm, $E D = 9$ cm and $B C = 20$ cm, find BF.
Hint
Apply BPT in $△ A D C$ (with $E G ∥ D C$ ) and in $△ C A B$ (with $GF ∥ A B$ ).
Answer
In $△ A D C$ : $\frac{A E}{E D} = \frac{A G}{GC}$ . In $△ C A B$ : $\frac{A G}{GC} = \frac{B F}{F C}$ . So $\frac{A E}{E D} = \frac{B F}{F C} = \frac{2}{3}$ , BF = $\frac{2}{5} \times 20 = 8$ cm.
Q3 · 3 marks · Short answerHarder board level
Using BPT, prove that a line drawn through the midpoint of one side of a triangle, parallel to another side, bisects the third side. Hence: in $△ A B C$ , D is the midpoint of AB, $D E ∥ B C$ meets AC at E, and $E F ∥ A B$ meets BC at F. If $B C = 14$ cm, find BF and DE.
Hint
With AD = DB, BPT gives AE/EC = 1. Apply the same result again in the second step.
Answer
If D is the midpoint of AB and $D E ∥ B C$ , BPT gives $\frac{A E}{E C} = \frac{A D}{D B} = 1$ , so $A E = E C$ : E bisects AC. Now E is the midpoint of AC and $E F ∥ A B$ , so by the same result F is the midpoint of BC and $B F = 7$ cm. Also $△ A D E \sim △ A B C$ (AA) with ratio $\frac{1}{2}$ , so $D E = 7$ cm (and BDEF is a parallelogram with $D E = B F$ ).
Q4 · 1 mark · MCQBoard level
In $△ A B C$ , $D E ∥ B C$ with D on AB and E on AC. Which of the following is NOT correct?
(a) $\frac{A D}{D B} = \frac{A E}{E C}$
(b) $\frac{A D}{A B} = \frac{A E}{A C}$
(c) $\frac{A D}{D B} = \frac{D E}{B C}$
(d) $\frac{D B}{A B} = \frac{E C}{A C}$
Hint
DE is compared with BC through the similar triangles ADE and ABC, using whole sides.
Answer
(c). (a) is BPT, and (b) and (d) follow from it. DE/BC equals AD/AB, not AD/DB.
Q5 · 5 marks · Long answerHarder board level
O is a point inside $△ A B C$ . Join OA, OB and OC. D, E and F lie on OA, OB and OC respectively, with $D E ∥ A B$ and $E F ∥ B C$ .
(i) Prove that $D F ∥ A C$ .
(ii) If $O D = 2$ cm, $D A = 3$ cm, $O E = 4$ cm and $O F = 3$ cm, find EB and FC, and verify the result of (i) with these lengths.
Hint
Use BPT in $△ O A B$ and $△ O B C$ , then the converse of BPT in $△ O A C$ .
Answer
(i) In $△ O A B$ , $D E ∥ A B$ , so by BPT $\frac{O D}{D A} = \frac{O E}{E B}$ . In $△ O B C$ , $E F ∥ B C$ , so $\frac{O E}{E B} = \frac{O F}{F C}$ . Hence $\frac{O D}{D A} = \frac{O F}{F C}$ , and by the converse of BPT in $△ O A C$ , $D F ∥ A C$ . (ii) $\frac{2}{3} = \frac{4}{E B}$ , so EB = 6 cm. $\frac{4}{6} = \frac{3}{F C}$ , so FC = 4.5 cm. Check: $\frac{O D}{D A} = \frac{2}{3}$ and $\frac{O F}{F C} = \frac{3}{4.5} = \frac{2}{3}$ , equal, so $D F ∥ A C$ .
Q6 · 5 marks · Long answerHarder board level
In $△ A B C$ , $∠ B = 9 0^{\circ}$ , $A B = 9$ cm and $B C = 12$ cm. D is on AB and E is on AC with $D E ∥ B C$ and $A D = 3$ cm. Using Pythagoras theorem as a known result, find AC, AE, DE the perimeter of the quadrilateral DBCE, and the ratio of the perimeters of $△ A D E$ and $△ A B C$ .
Hint
Find AC first. $∠ A D E = ∠ A B C$ (corresponding angles), so $△ A D E$ is also right-angled.
Answer
$A C = 81 + 144 = 15$ cm. By BPT $\frac{A D}{A B} = \frac{A E}{A C}$ : $\frac{3}{9} = \frac{A E}{15}$ , AE = 5 cm, so EC = 10 cm. $∠ A D E = 9 0^{\circ}$ , so $D E = 25 - 9 = 4$ cm (also $\frac{D E}{B C} = \frac{A D}{A B} = \frac{1}{3}$ gives 4 cm). DB = 6 cm. Perimeter of DBCE = 6 + 12 + 10 + 4 = 32 cm. Perimeter of $△ A D E = 3 + 4 + 5 = 12$ cm and of $△ A B C = 9 + 12 + 15 = 36$ cm, ratio $1 : 3$ , the same as the side ratio $A D : A B$ .
Q7 · 4 marks · Case-basedHOTS / competency
Suppose a triangular school garden ABC has a straight path DE laid parallel to the fence BC, with D on AB and E on AC. The gardener measures $A D = (x - 1)$ m, $D B = (x + 1)$ m, $A E = (x + 1)$ m and $E C = (2 x - 1)$ m.
(i) Write the equation in x given by BPT. (1)
(ii) Find x, giving a reason for rejecting any value. (1)
(iii) Find AB and AC, and the length of the path DE if BC = 20 m. (2)
Hint
For DE, use similarity with the whole side AB.
Answer
(i) $\frac{x - 1}{x + 1} = \frac{x + 1}{2 x - 1}$ , that is $(x - 1) (2 x - 1) = (x + 1)^{2}$ . (ii) $2 x^{2} - 3 x + 1 = x^{2} + 2 x + 1$ , so $x^{2} - 5 x = 0$ , x = 0 or 5. x = 0 makes AD = −1, impossible, so x = 5. (iii) AD = 4, DB = 6, AE = 6, EC = 9, so AB = 10 m and AC = 15 m. $△ A D E \sim △ A B C$ (AA), so $D E = \frac{4}{10} \times 20 = 8$ m.
Q8 · 1 mark · Assertion-reasonHarder board level
Assertion (A): In a trapezium ABCD with $A B ∥ D C$ , a line EF parallel to AB, with E on AD and F on BC, divides AD and BC in the same ratio.
Reason (R): Drawing the diagonal AC and applying BPT in $△ A D C$ and in $△ C A B$ gives $\frac{A E}{E D}$ and $\frac{B F}{F C}$ both equal to the ratio in which EF divides AC.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Let EF cut AC at G and write both BPT ratios.
Answer
(a). With G on AC: $\frac{A E}{E D} = \frac{A G}{GC}$ (BPT in $△ A D C$ , $E G ∥ D C$ ) and $\frac{B F}{F C} = \frac{A G}{GC}$ (BPT in $△ C A B$ , $GF ∥ A B$ ). So A is true, and R is exactly its proof.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State BPT.Show answer
A line parallel to one side of a triangle meeting the other two sides at distinct points divides them in the same ratio.
2.Another name for BPT?Show answer
Thales theorem.
3.Part-to-whole form of BPT?Show answer
$\frac{A D}{A B} = \frac{A E}{A C}$
4.Why is ar(BDE) = ar(DEC) in the proof?Show answer
Same base DE, between the same parallels DE and BC.
5.What construction is used in the proof?Show answer
Join BE and CD; draw $E N ⊥ A B$ and $D M ⊥ A C$ .

Basic Proportionality Theorem (Thales Theorem): frequently asked questions

Do I need to learn the proof of BPT?

Yes. It is the only theorem in the Triangles chapter that the syllabus asks you to prove. Practise writing it with given, to prove, construction and reasons.

Do I need to prove the converse of BPT?

No. The converse is stated without proof. You only quote and use it.

Can I use BPT to find the length of DE?

No. BPT only links the pieces of the two cut sides. For DE, prove the small and big triangles similar by AA and use the whole side ratio.

Does the BPT proof need any special area result?

No. The area step in the proof uses only the formula half × base × height, and the fact that triangles on the same base and between the same parallels have equal areas.