MathsCBSE Class 10 · Triangles

Converse of Basic Proportionality Theorem, Class 10

Updated 14 October 2026 · Fact-checked

The converse of the Basic Proportionality Theorem says that if a line divides two sides of a triangle in the same ratio, it is parallel to the third side. It is stated without proof. Board questions ask you to prove a line parallel from lengths, find an unknown that makes it parallel, then find lengths by similarity.

When AD/DB equals AE/EC (here 2/3 and 2/3), the line DE must be parallel to BC; unequal ratios would mean not parallel.

Illustration drawn for QuestionBuddy.

Understand Converse of Basic Proportionality Theorem

The Basic Proportionality Theorem (BPT) starts with a parallel line and gives you equal ratios: if $D E ∥ B C$ in $△ A B C$ , then $\frac{A D}{D B} = \frac{A E}{E C}$ . The converse runs the other way. It starts with equal ratios and gives you a parallel line.

Statement: If a line divides two sides of a triangle in the same ratio, then the line is parallel to the third side. So in $△ A B C$ , with D on AB and E on AC, if $\frac{A D}{D B} = \frac{A E}{E C}$ , then $D E ∥ B C$ . In the NCERT book this is Theorem 6.2, and the syllabus asks you to state and use it, not prove it.

Why is it believable? Through D there is exactly one line parallel to BC. By BPT, that line meets AC at a point E' with $\frac{A D}{D B} = \frac{A E ^{'}}{E ^{'} C}$ . If E also gives this same ratio, E and E' split AC in the same ratio, so they are the same point. Hence DE is that parallel line.

Conditions to check:

D and E must lie on two sides of the same triangle, both measured from the same vertex A.
Compare like with like: part to part ( $\frac{A D}{D B}$ with $\frac{A E}{E C}$ ) or part to whole ( $\frac{A D}{A B}$ with $\frac{A E}{A C}$ ). Never mix a part on one side with a whole on the other.
If the ratios are unequal, the line is not parallel to the third side.

The converse links to the rest of the chapter. Once $D E ∥ B C$ , corresponding angles give $∠ A D E = ∠ A B C$ and $∠ A E D = ∠ A C B$ , so $△ A D E \sim △ A B C$ by AA. That is how you find the third side: $\frac{D E}{B C} = \frac{A D}{A B}$ . It also lets you chain results: if two lines are each shown parallel to a third line, they are parallel to each other.

Converse of Basic Proportionality Theorem: formulas to remember

Converse of BPT (part to part): $\frac{A D}{D B} = \frac{A E}{E C} \Rightarrow D E ∥ B C$; D on AB, E on AC. Stated without proof (Theorem 6.2). Use to show a line is parallel.
Converse of BPT (part to whole): $\frac{A D}{A B} = \frac{A E}{A C} \Rightarrow D E ∥ B C$; Equivalent form. Handy when whole sides are given.
Test for not parallel: $\frac{A D}{D B} \neq = \frac{A E}{E C} \Rightarrow D E \neq ∥ B C$; If DE were parallel, BPT would force equal ratios.
Third side after proving parallel: $D E ∥ B C \Rightarrow △ A D E \sim △ A B C \Rightarrow \frac{D E}{B C} = \frac{A D}{A B}$; Uses AA similarity. Use the whole side AB, not DB.
Transitivity of parallel lines: $l ∥ m, n ∥ m \Rightarrow l ∥ n$; Used when the converse is applied in two triangles sharing a side.

How to solve Converse of Basic Proportionality Theorem questions

Use this method on any question that asks you to show a line is parallel or to find a value that makes it parallel.

1Name the triangle and confirm both points lie on its two sides, measured from the common vertex.
2Find any missing parts: part = whole minus other part.
3Write both ratios in the same form (part to part, or part to whole) and simplify each to lowest terms.
4If both ratios are equal, write: "Since $\frac{A D}{D B} = \frac{A E}{E C}$ , by the converse of BPT, $D E ∥ B C$ ."
5If an unknown is involved, set the ratios equal, cross-multiply, solve, and reject any value giving a zero or negative length.
6For a third side or angle, use corresponding angles from the parallel lines, state $△ A D E \sim △ A B C$ (AA), and use $\frac{D E}{B C} = \frac{A D}{A B}$ .
7Check by substituting your lengths back into both ratios and give units.

Converse of Basic Proportionality 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.

Converse of the Basic Proportionality Theorem

Imagine your school is painting a triangular banner ABC for Sports Day, with top corner A and base BC 120 cm long. A gold ribbon will be stitched from point D on side AB to point E on side AC. On the banner, AB = 150 cm and AC = 180 cm. The tailor marks AD = 100 cm and EC = 60 cm.

Will the ribbon DE be parallel to the base BC? If yes, find the length of ribbon needed. If the tailor instead marked EC = 50 cm, would it still be parallel?

Try it first, then show the solutionSolution

DB = 150 − 100 = 50 cm, so $\frac{A D}{D B} = \frac{100}{50} = 2$ .
AE = 180 − 60 = 120 cm, so $\frac{A E}{E C} = \frac{120}{60} = 2$ .
Ratios equal, so by the converse of BPT, $D E ∥ B C$ .
$∠ A D E = ∠ A B C$ (corresponding), $∠ A$ common, so $△ A D E \sim △ A B C$ (AA).
$\frac{D E}{120} = \frac{A D}{A B} = \frac{100}{150} = \frac{2}{3}$ , so DE = 80 cm.
With EC = 50 cm: AE = 130 cm, $\frac{A E}{E C} = \frac{13}{5} \neq = 2$ , so not parallel.

Answer: Yes, DE is parallel to BC and needs 80 cm of ribbon. With EC = 50 cm it would not be parallel.

Memory hook: Same split on both sides, the ribbon runs straight along the base.

Tricks to remember Converse of Basic Proportionality Theorem

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

Forward and back
BPT: parallel → ratio. Converse: ratio → parallel. Read the question: if parallel is given, use BPT; if you must show parallel, use the converse.
Use it when: Choosing which result to quote.
Like with like
Top over bottom on both sides, or top over whole on both sides. Never top over bottom against top over whole.
Use it when: Setting up the ratio correctly.
Start at the same corner
Every first length in each ratio starts from the shared vertex A.
Use it when: Avoids flipping a ratio by measuring from the wrong end.

Converse of Basic Proportionality Theorem: solved examples

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

Example 1: Checking for parallel with jumbled data · 2 marks · Very short answerBoard level

In $△ P QR$ , S and T are points on PQ and PR. $P S = 3$ cm, $S Q = 4.5$ cm, $P T = 4$ cm and $P R = 10$ cm. Is $S T ∥ QR$ ? Justify.

Try it first, then show the solutionSolution

TR = PR − PT = 10 − 4 = 6 cm. (½)
$\frac{P S}{S Q} = \frac{3}{4.5} = \frac{2}{3}$ . (½)
$\frac{P T}{T R} = \frac{4}{6} = \frac{2}{3}$ . (½)
Ratios are equal, so by the converse of BPT, $S T ∥ QR$ . (½)

Answer: Yes, $S T ∥ QR$ since $\frac{P S}{S Q} = \frac{P T}{T R} = \frac{2}{3}$ .

Example 2: Finding x for a parallel line · 3 marks · Short answerBoard level

In $△ A B C$ , D is on AB and E is on AC with $A D = (x + 1)$ cm, $D B = (3 x - 1)$ cm, $A E = (x + 3)$ cm and $E C = (3 x + 4)$ cm. Find the value of $x$ for which $D E ∥ B C$ , and find AB and AC.

Try it first, then show the solutionSolution

For $D E ∥ B C$ , by the converse of BPT we need $\frac{x + 1}{3 x - 1} = \frac{x + 3}{3 x + 4}$ . (½)
$(x + 1) (3 x + 4) = (x + 3) (3 x - 1)$ . (½)
$3 x^{2} + 7 x + 4 = 3 x^{2} + 8 x - 3$ . (½)
$x = 7$ . (½)
AD = 8, DB = 20, so AB = 28 cm; AE = 10, EC = 25, so AC = 35 cm. (½)
Check: $\frac{8}{20} = \frac{10}{25} = \frac{2}{5}$ . (½)

Answer: $x = 7$ ; AB = 28 cm, AC = 35 cm.

Example 3: Show parallel, then find the third side · 3 marks · Short answerBoard level

In $△ L M N$ , X is on LM and Y is on LN. $L M = 12$ cm, $X M = 4.5$ cm, $L N = 16$ cm and $L Y = 10$ cm. Show that $X Y ∥ M N$ . If $M N = 20$ cm, find XY and the perimeter of $△ L X Y$ .

Try it first, then show the solutionSolution

LX = 12 − 4.5 = 7.5 cm, YN = 16 − 10 = 6 cm. (½)
$\frac{L X}{X M} = \frac{7.5}{4.5} = \frac{5}{3}$ and $\frac{L Y}{Y N} = \frac{10}{6} = \frac{5}{3}$ . (½)
By the converse of BPT, $X Y ∥ M N$ . (½)
$∠ L X Y = ∠ L M N$ (corresponding), $∠ L$ common, so $△ L X Y \sim △ L M N$ (AA). (½)
$\frac{X Y}{20} = \frac{7.5}{12} = \frac{5}{8}$ , so XY = 12.5 cm. (½)
Perimeter = 7.5 + 10 + 12.5 = 30 cm. (½)

Answer: $X Y ∥ M N$ ; XY = 12.5 cm; perimeter of $△ L X Y$ = 30 cm.

Example 4: Converse used twice in a quadrilateral · 5 marks · Long answerHarder board level

ABCD is a quadrilateral and O is a point inside it. P, Q, R are points on OA, OB, OC respectively such that $P Q ∥ A B$ and $QR ∥ B C$ . Prove that $P R ∥ A C$ . If $O P = 2$ cm, $P A = 3$ cm and $A C = 15$ cm, find PR.

Try it first, then show the solutionSolution

In $△ O A B$ , $P Q ∥ A B$ , so by BPT $\frac{O P}{P A} = \frac{O Q}{QB}$ ... (1). (1)
In $△ O B C$ , $QR ∥ B C$ , so by BPT $\frac{O Q}{QB} = \frac{O R}{R C}$ ... (2). (1)
From (1) and (2), $\frac{O P}{P A} = \frac{O R}{R C}$ . (½)
In $△ O A C$ , P and R divide OA and OC in the same ratio, so by the converse of BPT, $P R ∥ A C$ . (1)
$∠ O P R = ∠ O A C$ (corresponding), $∠ O$ common, so $△ O P R \sim △ O A C$ (AA). (½)
$\frac{P R}{A C} = \frac{O P}{O A} = \frac{2}{5}$ . (½)
PR = $\frac{2}{5} \times 15$ = 6 cm. (½)

Answer: $P R ∥ A C$ (proved); PR = 6 cm.

Example 5: Case: the garden trellis · 4 marks · Case-basedHOTS / competency

Suppose a gardener fixes a triangular trellis ABC against a wall. Side AB runs straight up the wall with A at the top and B on the ground, side BC runs along level ground, and $∠ A B C = 9 0^{\circ}$ . AB = 1.8 m and BC = 2.4 m; the sloping side AC has not been measured. Wooden strip DE joins D on AB to E on AC, with AD = 0.6 m and AE = 1 m. A second strip FG joins F on AB to G on AC, with BF = 0.6 m and CG = 1.2 m.
(i) Find the length of the sloping side AC. (1)
(ii) Is DE parallel to BC? Give a reason. (1)
(iii) Find the perimeter of $△ A D E$ . Is FG parallel to BC? Give a reason. (2)

Try it first, then show the solutionSolution

(i) Using Pythagoras as a known result in right $△ A B C$ : $A C^{2} = 1. 8^{2} + 2. 4^{2} = 3.24 + 5.76 = 9$ , so AC = 3 m. (1)
(ii) $\frac{A D}{A B} = \frac{0.6}{1.8} = \frac{1}{3}$ and $\frac{A E}{A C} = \frac{1}{3}$ ; equal, so DE ∥ BC by the converse of BPT. (1)
(iii) $∠ A D E = ∠ A B C$ (corresponding), $∠ A$ common, so $△ A D E \sim △ A B C$ (AA) with ratio $\frac{1}{3}$ . Perimeter of $△ A B C$ = 1.8 + 2.4 + 3 = 7.2 m, so perimeter of $△ A D E$ = $\frac{1}{3} \times 7.2$ = 2.4 m. (1)
AF = 1.8 − 0.6 = 1.2 m, $\frac{A F}{A B} = \frac{2}{3}$ ; AG = 3 − 1.2 = 1.8 m, $\frac{A G}{A C} = \frac{3}{5}$ ; unequal, so FG is not parallel to BC. (1)

Answer: (i) AC = 3 m. (ii) Yes, DE ∥ BC. (iii) Perimeter of $△ A D E$ = 2.4 m; FG is not parallel to BC.

Example 6: Assertion-reason: which result explains it? · 1 mark · Assertion-reasonHarder board level

Assertion (A): In $△ A B C$ , D is on AB and E is on AC with AD = 2 cm, DB = 3 cm, AE = 3 cm and AC = 7.5 cm. Then $D E ∥ B C$ .
Reason (R): 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.
(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

EC = 7.5 − 3 = 4.5 cm, so $\frac{A E}{E C} = \frac{3}{4.5} = \frac{2}{3} = \frac{A D}{D B}$ ; by the converse of BPT, $D E ∥ B C$ , so A is true. (½)
R is BPT, which is true, but it goes from parallel to ratio; A needs ratio to parallel, which is the converse. So R does not explain A. (½)

Answer: (b) Both A and R are true but R is not the correct explanation of A.

Example 7: Quadratic in x with a root to reject · 5 marks · Long answerHarder board level

In $△ A B C$ , D is on AB and E is on AC with $A D = (x + 2)$ cm, $D B = (x - 1)$ cm, $A E = (3 x - 4)$ cm and $E C = x$ cm. Find the value of $x$ for which $D E ∥ B C$ . If $B C = 15$ cm, find DE and the perimeter of the trapezium DBCE.

Try it first, then show the solutionSolution

For $D E ∥ B C$ , by the converse of BPT we need $\frac{x + 2}{x - 1} = \frac{3 x - 4}{x}$ . (½)
Cross-multiplying: $x (x + 2) = (x - 1) (3 x - 4)$ . (½)
$x^{2} + 2 x = 3 x^{2} - 7 x + 4$ , so $2 x^{2} - 9 x + 4 = 0$ . (1)
$(2 x - 1) (x - 4) = 0$ , so $x = \frac{1}{2}$ or $x = 4$ . (½)
$x = \frac{1}{2}$ gives $D B = - \frac{1}{2}$ cm, which is impossible, so $x = 4$ . (½)
AD = 6, DB = 3, AE = 8, EC = 4; check $\frac{6}{3} = \frac{8}{4} = 2$ . AB = 9 cm. (½)
$∠ A D E = ∠ A B C$ (corresponding), $∠ A$ common, so $△ A D E \sim △ A B C$ (AA); $\frac{D E}{15} = \frac{A D}{A B} = \frac{6}{9}$ , so DE = 10 cm. (1)
Perimeter of DBCE = DB + BC + CE + ED = 3 + 15 + 4 + 10 = 32 cm. (½)

Answer: $x = 4$ (reject $x = \frac{1}{2}$ ); DE = 10 cm; perimeter of DBCE = 32 cm.

Example 8: Proving a quadrilateral is a trapezium · 5 marks · Long answerHarder board level

In quadrilateral ABCD, the diagonals AC and BD intersect at O such that $\frac{A O}{O C} = \frac{B O}{O D}$ . Prove that $A B ∥ D C$ , so ABCD is a trapezium. If AB = 18 cm, AO = 6 cm and OC = 4 cm, find DC.

Try it first, then show the solutionSolution

Construction: through O draw $O E ∥ D C$ meeting AD at E. (½)
In $△ A D C$ , $O E ∥ D C$ , so by BPT $\frac{A E}{E D} = \frac{A O}{O C}$ ... (1). (1)
Given $\frac{A O}{O C} = \frac{B O}{O D}$ , so from (1), $\frac{A E}{E D} = \frac{B O}{O D}$ ... (2). (½)
In $△ D A B$ , E on DA and O on DB divide the sides in the same ratio by (2), so by the converse of BPT, $E O ∥ A B$ . (1)
$E O ∥ A B$ and $E O ∥ D C$ , so $A B ∥ D C$ ; hence ABCD is a trapezium. (½)
$∠ O A B = ∠ O C D$ (alternate angles, $A B ∥ D C$ ) and $∠ A O B = ∠ C O D$ (vertically opposite), so $△ A O B \sim △ C O D$ (AA). (1)
$\frac{A B}{C D} = \frac{A O}{O C} = \frac{6}{4}$ , so DC = $\frac{4}{6} \times 18$ = 12 cm. (½)

Answer: $A B ∥ D C$ (proved), so ABCD is a trapezium; DC = 12 cm.

Where students lose marks in Converse of Basic Proportionality Theorem

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

Mistake: Mixing part-to-part with part-to-whole
The two ratios measure different things, so they look unequal even when DE is parallel.
Wrong working
$\frac{A D}{A B} = \frac{3}{9}$ , $\frac{A E}{E C} = \frac{4}{8}$ , unequal, so not parallel.
Right working
DB = 6: $\frac{A D}{D B} = \frac{3}{6} = \frac{1}{2} = \frac{4}{8} = \frac{A E}{E C}$ , so $D E ∥ B C$ .
Do this: Write both ratios in the same form before comparing.
Usually costs: Full marks for the part, as the conclusion is wrong.
Mistake: Quoting BPT instead of its converse
Students write "by BPT" when proving lines parallel; BPT assumes parallel lines.
Wrong working
$\frac{A D}{D B} = \frac{A E}{E C}$ , hence by BPT, $D E ∥ B C$ .
Right working
$\frac{A D}{D B} = \frac{A E}{E C}$ , hence by the converse of BPT, $D E ∥ B C$ .
Do this: Ratio to parallel is always the converse.
Usually costs: ½ to 1 mark for the reason.
Mistake: Using DB instead of AB for the third side
DE is found from similarity of $△ A D E$ and $△ A B C$ , whose sides are AD and AB.
Wrong working
$\frac{D E}{B C} = \frac{A D}{D B} = \frac{2}{3}$ .
Right working
$\frac{D E}{B C} = \frac{A D}{A B} = \frac{2}{5}$ .
Do this: Third side needs part over whole.
Usually costs: 1 to 2 marks in a 3-mark question.
Mistake: Not rejecting an invalid root
A value of x can make a length zero or negative, even when x itself is positive.
Wrong working
With $D B = (x - 1)$ cm: $x = 4$ or $x = \frac{1}{2}$ ; both accepted.
Right working
$x = \frac{1}{2}$ makes $D B = - \frac{1}{2}$ cm, so only $x = 4$ .
Do this: Substitute each root into every length.
Usually costs: ½ mark.

Exam tips

Always write the reason in words: "by the converse of BPT". Stating the reason in words usually earns a mark in the marking scheme, so always write it.
Show the missing segment calculation and both simplified ratios; a bare "equal, so parallel" can lose marks.
When asked for the third side, first prove $△ A D E \sim △ A B C$ by AA using the parallel lines.
In assertion-reason items, check whether the reason is BPT or its converse and whether it actually explains the assertion.
In proofs inside quadrilaterals, combine BPT results from two triangles, then apply the converse in the third.

Practice questions on Converse of Basic Proportionality Theorem

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

Q1 · 5 marks · Long answerHarder board level
In $△ A B C$ , D is on AB and E is on AC with $A D = 6$ cm, $D B = 9$ cm, $A E = 8$ cm, $E C = 12$ cm and $B C = 25$ cm. Show that $D E ∥ B C$ and find DE. F is a point on BC with $B F = 10$ cm; show that $E F ∥ A B$ . Hence show that DBFE is a parallelogram and find its perimeter.
Hint
For EF, use the converse in triangle CAB with points E on CA and F on CB, measured from C. Find CF first.
Answer
$\frac{6}{9} = \frac{8}{12} = \frac{2}{3}$ , so $D E ∥ B C$ by the converse of BPT. $△ A D E \sim △ A B C$ (AA), so DE = $\frac{6}{15} \times 25$ = 10 cm. CF = 25 − 10 = 15 cm; $\frac{C E}{E A} = \frac{12}{8} = \frac{3}{2}$ and $\frac{C F}{F B} = \frac{15}{10} = \frac{3}{2}$ . Equal, so $E F ∥ A B$ by the converse of BPT. Now $D E ∥ B F$ and $E F ∥ D B$ , so DBFE is a parallelogram. $△ C E F \sim △ C A B$ (AA), so $E F = \frac{12}{20} \times 15 = 9$ cm $= D B$ , and $B F = 10$ cm $= D E$ , as a parallelogram needs. Perimeter $= 2 (10 + 9) = 38$ cm.
Q2 · 5 marks · Long answerHarder board level
O is any point inside $△ P QR$ . Points X, Y, Z lie on OP, OQ, OR with $\frac{O X}{X P} = \frac{O Y}{Y Q} = \frac{O Z}{Z R} = \frac{1}{2}$ . Prove that $X Y ∥ P Q$ , $Y Z ∥ QR$ , and find the perimeter of $△ X Y Z$ if the perimeter of $△ P QR$ is 36 cm.
Hint
Apply the converse in triangles OPQ and OQR, then AA similarity gives each side as one-third.
Answer
In $△ O P Q$ , $\frac{O X}{X P} = \frac{O Y}{Y Q}$ , so $X Y ∥ P Q$ (converse of BPT); likewise $Y Z ∥ QR$ and $X Z ∥ P R$ . $△ O X Y \sim △ O P Q$ gives $X Y = \frac{1}{3} P Q$ ; similarly for the others. Perimeter = $\frac{1}{3} \times 36$ = 12 cm.
Q3 · 2 marks · Very short answerBoard level
Aman says: "In $△ A B C$ , D is on AB and E is on AC with AD = 4 cm, AB = 10 cm, AE = 3 cm and EC = 4.5 cm. Since $\frac{A D}{A B} = \frac{2}{5}$ and $\frac{A E}{E C} = \frac{2}{3}$ are unequal, DE is not parallel to BC." Find his error and give the correct conclusion.
Hint
Check which kind of ratio he compared on each side.
Answer
He compared part-to-whole with part-to-part. Correct: DB = 6, $\frac{A D}{D B} = \frac{4}{6} = \frac{2}{3} = \frac{A E}{E C}$ . So $D E ∥ B C$ by the converse of BPT.
Q4 · 1 mark · MCQBoard level
In $△ A B C$ , D is on AB, E is on BC and F is on CA, with AD = 3 cm, DB = 6 cm, BE = 4 cm, EC = 4 cm, AF = 5 cm and FC = 10 cm. Which of the following is true?
(a) $D E ∥ A C$
(b) $E F ∥ A B$
(c) $D F ∥ B C$
(d) None of these
Hint
Test each line from the vertex it is opposite to: B for DE, C for EF, A for DF.
Answer
(c) $D F ∥ B C$ . From A: $\frac{A D}{D B} = \frac{3}{6} = \frac{1}{2}$ and $\frac{A F}{F C} = \frac{5}{10} = \frac{1}{2}$ , equal. From B: $\frac{B D}{D A} = 2$ but $\frac{B E}{E C} = 1$ ; from C: $\frac{C E}{E B} = 1$ but $\frac{C F}{F A} = 2$ . So only DF is parallel.
Q5 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a school banner is a triangle ABC with BC = 7.5 m along the bottom. A strip of gold tape DE is to run from D on AB to E on AC. The tailor's notes say $A D = x$ m, $D B = 2$ m, $A E = (x + 2)$ m and $E C = (x - 1)$ m, and that DE must be parallel to BC.
(i) Write the condition on x given by the converse of BPT. (1 mark)
(ii) Find x, giving a reason for rejecting any value. (1 mark)
(iii) Find DE. Edging tape is sewn round the top piece ADE and round the strip DBCE separately. Which piece needs more tape, and by how much? (2 marks)
Hint
The ratio equation is a quadratic. After finding x, use AA to get DE.
Answer
(i) $\frac{x}{2} = \frac{x + 2}{x - 1}$ . (1)
(ii) $x (x - 1) = 2 (x + 2)$ , so $x^{2} - 3 x - 4 = 0$ and $(x - 4) (x + 1) = 0$ . $x = - 1$ is not a length, so $x = 4$ : AD = 4, AE = 6, EC = 3, so AB = 6 m and AC = 9 m. (1)
(iii) $D E ∥ B C$ , so $△ A D E \sim △ A B C$ (AA) and $D E = \frac{4}{6} \times 7.5 = 5$ m. Perimeter of ADE $= 4 + 6 + 5 = 15$ m; perimeter of DBCE $= 2 + 7.5 + 3 + 5 = 17.5$ m. The strip DBCE needs 2.5 m more tape. (2)
Q6 · 2 marks · Very short answerBoard level
In $△ A B C$ , D and E are the midpoints of AB and AC. Using the converse of BPT, show that $D E ∥ B C$ .
Hint
Midpoints make both ratios equal to 1.
Answer
D is the midpoint of AB, so $A D = D B$ and $\frac{A D}{D B} = 1$ (½ mark). E is the midpoint of AC, so $\frac{A E}{E C} = 1$ (½ mark). The ratios are equal, so by the converse of BPT, $D E ∥ B C$ (1 mark).
Q7 · 5 marks · Long answerHOTS / competency
ABCD is a quadrilateral with diagonal BD = 20 cm. P and Q lie on AB and AD with $\frac{A P}{P B} = \frac{A Q}{Q D} = \frac{3}{2}$ . R and S lie on CB and CD with $\frac{C R}{R B} = \frac{C S}{S D} = \frac{3}{2}$ . Prove that $P Q ∥ R S$ , find PQ and RS, and hence show that PQRS is a parallelogram.
Hint
Use the diagonal BD as the third side of two different triangles.
Answer
In $△ A B D$ , $\frac{A P}{P B} = \frac{A Q}{Q D}$ , so $P Q ∥ B D$ by the converse of BPT. In $△ C B D$ , $\frac{C R}{R B} = \frac{C S}{S D}$ , so $R S ∥ B D$ . Hence $P Q ∥ R S$ (both parallel to BD). $△ A P Q \sim △ A B D$ (AA), so $P Q = \frac{A P}{A B} \times B D = \frac{3}{5} \times 20 = 12$ cm. Likewise $△ C R S \sim △ C B D$ , so $R S = \frac{3}{5} \times 20 = 12$ cm. One pair of opposite sides, PQ and RS, is both parallel and equal, so PQRS is a parallelogram.

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1.State the converse of BPT.Show answer
If a line divides two sides of a triangle in the same ratio, it is parallel to the third side.
2.Is the converse of BPT proved in the 2026-27 syllabus?Show answer
No, it is stated without proof.
3.Part-to-whole form of the converse?Show answer
$\frac{A D}{A B} = \frac{A E}{A C} \Rightarrow D E ∥ B C$ .
4.If $\frac{A D}{D B} \neq = \frac{A E}{E C}$ , what follows?Show answer
DE is not parallel to BC.
5.Which criterion gives DE after proving parallel?Show answer
AA similarity: $△ A D E \sim △ A B C$ .

Converse of Basic Proportionality Theorem: frequently asked questions

What is the difference between BPT and its converse?

BPT starts with a line parallel to one side and concludes the other two sides are divided in the same ratio. The converse starts with the equal ratio and concludes the line is parallel. Use the one that matches what is given.

Do I need to prove the converse of BPT in the board exam?

No. The syllabus says it is stated without proof. You must state it correctly and use it, but only BPT itself needs a full proof.

Can I compare AD over AB with AE over AC?

Yes, part-to-whole on both sides is a valid form. Just do not compare part-to-whole on one side with part-to-part on the other.

Is the area of similar triangles theorem used here?

No. The theorem on areas of similar triangles is not in the 2026-27 syllabus.