Mid-point and Intercept Theorems ICSE Class 9: Proofs and Applications

What Mid-point and Intercept Theorems is about

Weightage: CISCE publishes no chapter weightage and no question-paper layout for ICSE Class 9 Mathematics. Class IX is examined by your school: one 80-mark paper of 3 hours plus 20 marks of internal assessment. Mid-point and Intercept Theorems is in the 4. Geometry (i) Triangles (c) part of the 2028 syllabus.Source: CISCE ICSE 2028 Mathematics syllabus, Class IX

Key concepts

The mid-point theorem

Theorem: the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.

In $△ABC$ with $D$ the mid-point of $AB$ and $E$ the mid-point of $AC$: $DE\parallel BC$ and $DE=\frac{1}{2}BC$.

Proof outline: produce $DE$ to $F$ so that $EF=DE$ and join $CF$. Then $△ADE\cong △CFE$ (SAS: $AE=CE$, $DE=FE$, vertically opposite angles at $E$). So $CF=AD=BD$ and $\angle ADE=\angle CFE$, which are alternate angles, so $CF\parallel BD$. Hence $BDFC$ is a parallelogram (one pair of opposite sides equal and parallel), so $DF\parallel BC$ and $DF=BC$. As $DE=\frac{1}{2}DF$, we get $DE=\frac{1}{2}BC$.

Worked example

In $△ABC$, $D$ and $E$ are mid-points of $AB$ and $AC$. If $BC=12$ cm and $\angle ABC=65^{\circ }$, find $DE$ and $\angle ADE$.

Try it first, then show the solutionSolution

1. $DE=\frac{1}{2}BC=6$ cm.
2. $DE\parallel BC$, so $\angle ADE=\angle ABC=65^{\circ }$ (corresponding angles).

Answer: DE = 6 cm; ∠ADE = 65°

The converse of the mid-point theorem

Theorem: the line drawn through the mid-point of one side of a triangle, parallel to another side, bisects the third side.

In $△ABC$, if $D$ is the mid-point of $AB$ and $DE\parallel BC$ meets $AC$ at $E$, then $E$ is the mid-point of $AC$.

Proof outline: through $C$ draw $CF\parallel AB$ to meet $DE$ produced at $F$. Then $BDFC$ is a parallelogram (both pairs of opposite sides parallel), so $CF=BD=AD$. In $△ADE$ and $△CFE$: $AD=CF$, $\angle DAE=\angle FCE$ (alternate), $\angle ADE=\angle CFE$ (alternate). So the triangles are congruent (AAS) and $AE=CE$.

Worked example

In $△PQR$, $M$ is the mid-point of $PQ$ and a line through $M$ parallel to $QR$ meets $PR$ at $N$. If $PR=9$ cm and $QR=11$ cm, find $PN$ and $MN$.

Try it first, then show the solutionSolution

1. By the converse, $N$ is the mid-point of $PR$: $PN=4.5$ cm.
2. Now $M$, $N$ are mid-points, so $MN=\frac{1}{2}QR=5.5$ cm.

Answer: PN = 4.5 cm; MN = 5.5 cm

The equal intercept theorem

A transversal is a line cutting two or more lines; the piece of it between two of those lines is an intercept.

Theorem: if a transversal makes equal intercepts on three (or more) parallel lines, then any other transversal cutting them also makes equal intercepts.

If lines $l\parallel m\parallel n$ cut transversal $p$ at $A$, $B$, $C$ with $AB=BC$, and transversal $q$ at $D$, $E$, $F$, then $DE=EF$.

Proof outline: join $AF$, meeting $m$ at $G$. In $△ACF$, $B$ is the mid-point of $AC$ and $BG\parallel CF$, so $G$ is the mid-point of $AF$ (converse). In $△FAD$, $G$ is the mid-point of $AF$ and $GE\parallel AD$, so $E$ is the mid-point of $DF$ (converse). So $DE=EF$.

Worked example

Three parallel lines cut transversal $p$ at $A$, $B$, $C$ with $AB=BC=3$ cm, and transversal $q$ at $D$, $E$, $F$ with $DE=4$ cm. Find $DF$.

Try it first, then show the solutionSolution

1. Equal intercepts on $p$, so equal intercepts on $q$: $EF=DE=4$ cm.
2. $DF=4+4=8$ cm.

Answer: 8 cm

Triangle of mid-points

Join the three mid-points $D$, $E$, $F$ of the sides of $△ABC$. By the mid-point theorem, each side of $△DEF$ is parallel to one side of $△ABC$ and half its length. So:

• the perimeter of $△DEF$ is half the perimeter of $△ABC$;
• the four small triangles formed ($△ADF$, $△DBE$, $△FEC$, $△EFD$) are congruent (SSS);
• each of $ADEF$, $DBEF$, $DECF$ is a parallelogram.

Worked example

The sides of $△ABC$ are 8 cm, 10 cm and 12 cm. Find the perimeter of the triangle formed by joining the mid-points of its sides.

Try it first, then show the solutionSolution

1. Each side of the new triangle is half a side of $△ABC$: 4, 5 and 6 cm.
2. Perimeter $=15$ cm (half of 30 cm).

Answer: 15 cm

Mid-points of a quadrilateral and a trapezium

Quadrilateral: joining the mid-points of the sides of any quadrilateral, in order, gives a parallelogram. Each pair of opposite sides is parallel to a diagonal and half its length, so its perimeter equals the sum of the diagonals.

Trapezium: in trapezium $ABCD$ with $AB\parallel DC$, the line through the mid-point $E$ of $AD$ parallel to $AB$ meets $BC$ at its mid-point $F$, and $EF=\frac{1}{2}(AB+DC)$. To prove this, join $AC$ meeting $EF$ at $G$ and apply the mid-point theorem in $△ADC$ and $△CAB$.

Worked example

In trapezium $ABCD$, $AB\parallel DC$, $AB=16$ cm and $DC=10$ cm. $E$ is the mid-point of $AD$ and $EF\parallel AB$ meets $BC$ at $F$. Find $EF$.

Try it first, then show the solutionSolution

1. Join $AC$ meeting $EF$ at $G$. In $△ADC$: $EG=\frac{1}{2}DC=5$ cm.
2. $G$ is the mid-point of $AC$ (converse), so in $△CAB$: $GF=\frac{1}{2}AB=8$ cm.
3. $EF=5+8=13$ cm.

Answer: 13 cm

Median to the hypotenuse

In a right-angled triangle, the mid-point of the hypotenuse is equidistant from all three vertices, so the median to the hypotenuse is half the hypotenuse.

Reason: in $△ABC$ with $\angle B=90^{\circ }$, let $M$ be the mid-point of $AC$. Draw $MN\parallel AB$ meeting $BC$ at $N$. By the converse, $N$ is the mid-point of $BC$; and $MN\perp BC$ (since $AB\perp BC$). Then $△MNB\cong △MNC$ (SAS), so $MB=MC=\frac{1}{2}AC$.

Worked example

A right-angled triangle has legs 5 cm and 12 cm. Find the length of the median to the hypotenuse.

Try it first, then show the solutionSolution

1. Hypotenuse $=\sqrt{25+144}=13$ cm.
2. Median $=\frac{13}{2}=6.5$ cm.

Answer: 6.5 cm

Maths in real life: Mid-point and Intercept Theorems

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.

Topics in this chapter

• Equal Intercepts, Trapezium Mid-points, Mid-point Quadrilateral and the Median to the Hypotenuse
• The Mid-point Theorem, Its Converse and the Triangle of Mid-points

Mid-point and Intercept Theorems formula sheet

Mid-point theorem

$DE\parallel BC,DE=\frac{1}{2}BC$

D, E mid-points of AB, AC.

Converse

$AD=DB,DE\parallel BC\Rightarrow AE=EC$

Line through a mid-point, parallel to a side.

Equal intercepts

$AB=BC\Rightarrow DE=EF$

On three parallel lines, for any transversal.

Mid-point triangle

$P_{DEF}=\frac{1}{2}{P}_{ABC}$

Four congruent small triangles.

Trapezium mid-line

$EF=\frac{1}{2}(AB+DC)$

E, F mid-points of the non-parallel sides.

Right-triangle median

$BM=\frac{1}{2}AC$

M mid-point of hypotenuse AC.

Mid-point and Intercept Theorems: diagrams to practise

Practise each one on paper: draw it from the checklist, then tick off every label.

How the school exam asks Mid-point and Intercept Theorems

The question patterns to expect, and the method that works for each.

Tricks to remember Mid-point and Intercept Theorems

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

• Half and parallel

  Mid-points joined: half as long, always parallel.

  Use it when: Mid-point theorem.

• Parallel from a middle hits a middle

  Start at a mid-point, go parallel, and you land on a mid-point.

  Use it when: Converse.

• Equal on one, equal on all

  Parallel lines that cut one transversal equally cut every transversal equally.

  Use it when: Equal intercept theorem.

• Average of the parallels

  Trapezium mid-line = average of the two parallel sides.

  Use it when: EF = ½(AB + DC).

Mid-point and Intercept Theorems: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 10-mark question split into 3 + 3 + 4 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Mid-point and Intercept Theorems

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

• ✗Mistake: Missing construction in a proof

  The proof of the mid-point theorem depends on producing $DE$ to $F$; without stating it, later steps have no basis.

  ✓Do this: Write 'Construction: …' as a separate line before the proof.

  Usually costs: 1 mark.

• ✗Mistake: Using the theorem when only one point is a mid-point

  The mid-point theorem needs two mid-points. With one mid-point and a parallel line you need the converse first.

  ✓Do this: Ask: do I know both mid-points? If not, use the converse to get the second one.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Doubling instead of halving (or the reverse)

  Students write $DE=2BC$.

  ✓Do this: The segment joining mid-points is the shorter one: always half.

  Usually costs: The answer.

• ✗Mistake: Wrong transversal in the intercept theorem

  The theorem needs the lines cut to be parallel; it says nothing if they are not.

  ✓Do this: State 'since l ∥ m ∥ n' before quoting equal intercepts.

  Usually costs: 1 mark.

• ✗Mistake: No reasons for parallel or equal statements

  Each step should cite the theorem, converse, or parallel-line angle property used.

  ✓Do this: Bracket the reason after each line.

  Usually costs: Method marks.

Practice questions on Mid-point and Intercept Theorems

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

1. Q1 · 1 mark · MCQExam level

   The figure formed by joining the mid-points of the sides of a rectangle, in order, is a:
   (a) rectangle
   (b) rhombus
   (c) square
   (d) trapezium

   Hint

   The diagonals of a rectangle are equal.

   Answer

   (b): all four sides equal half a diagonal.

2. Q2 · 2 marks · Short answerExam level

   In $△ABC$, $D$ and $E$ are mid-points of $AB$ and $AC$. If $DE=(x+3)$ cm and $BC=(3x-1)$ cm, find $x$ and $BC$.

   Hint

   BC = 2DE.

   Answer

   $3x-1=2x+6$, so $x=7$; $BC=20$ cm.

3. Q3 · 2 marks · Short answerExam level

   Three parallel lines cut one transversal at $A$, $B$, $C$ with $AB=BC$, and a second transversal at $P$, $Q$, $R$. If $PR=11$ cm, find $PQ$.

   Hint

   Equal intercepts.

   Answer

   $PQ=QR=5.5$ cm.

4. Q4 · 1 mark · Assertion-reasonHarder exam level

   Assertion (A): In $△ABC$, if $D$ is the mid-point of $AB$ and $DE\parallel BC$ meets $AC$ at $E$, then $DE=\frac{1}{2}BC$.
   Reason (R): The line through the mid-point of one side parallel to another side bisects the third side.
   (a) (A) is true and (R) is false.
   (b) (A) is false and (R) is true.
   (c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
   (d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).

   Hint

   R shows E is a mid-point; then what gives DE = ½BC?

   Answer

   (d). Both are true; (R) shows $E$ is a mid-point, but the length $DE=\frac{1}{2}BC$ comes from the mid-point theorem, so (R) alone does not explain (A).

5. Q5 · 3 marks · Short answerHarder exam level

   $ABCD$ is a parallelogram. $E$ is the mid-point of $AB$ and $F$ is the mid-point of $CD$. Prove that $AECF$ is a parallelogram.

   Hint

   AE and CF are halves of equal parallel sides.

   Answer

   $AE=\frac{1}{2}AB=\frac{1}{2}CD=CF$ and $AE\parallel CF$ (parts of $AB\parallel CD$). One pair of opposite sides equal and parallel, so $AECF$ is a parallelogram.

6. Q6 · 3 marks · Short answerHarder exam level

   In trapezium $PQRS$, $PQ\parallel SR$, $PQ=18$ cm and the mid-line joining the mid-points of $PS$ and $QR$ is 14 cm. Find $SR$.

   Hint

   Mid-line = ½(PQ + SR).

   Answer

   $14=\frac{1}{2}(18+SR)$, so $SR=10$ cm.

7. Q7 · 3 marks · Short answerHarder exam level

   In $△ABC$, $AD$ is a median and $E$ is the mid-point of $AD$. A line through $E$ parallel to $BC$ meets $AB$ at $P$ and $AC$ at $Q$. If $BC=16$ cm, find $PQ$.

   Hint

   Apply the converse in ΔABD and ΔADC.

   Answer

   In $△ABD$, $E$ is the mid-point of $AD$ and $PE\parallel BD$, so $PE=\frac{1}{2}BD=4$ cm. Similarly $EQ=\frac{1}{2}DC=4$ cm. $PQ=8$ cm.

8. Q8 · 4 marks · Long answerHarder exam level

   Prove that the line segments joining the mid-points of opposite sides of a quadrilateral bisect each other.

   Hint

   They are the diagonals of the mid-point parallelogram.

   Answer

   The mid-points $P$, $Q$, $R$, $S$ of $AB$, $BC$, $CD$, $DA$ form a parallelogram $PQRS$. $PR$ and $QS$ join mid-points of opposite sides and are its diagonals, and the diagonals of a parallelogram bisect each other.

9. Q9 · 3 marks · Short answerHarder exam level

   $ABC$ is a triangle right-angled at $B$. $D$ is the mid-point of $AC$ and $\angle ACB=32^{\circ }$. Find $\angle BDC$ and $\angle ADB$.

   Hint

   DB = DC.

   Answer

   $DB=DC$, so $\angle DBC=32^{\circ }$. $\angle BDC=180^{\circ }-64^{\circ }=116^{\circ }$; $\angle ADB=64^{\circ }$.

10. Q10 · 4 marks · Long answerHOTS / competency

    Using the equal intercept theorem, describe how to divide a 10 cm line segment $AB$ into three equal parts with a ruler and compasses, and justify it.

    Hint

    Draw a ray from A and mark three equal steps.

    Answer

    Draw any ray $AX$ at an acute angle to $AB$. With compasses mark $A_1$, $A_2$, $A_3$ on $AX$ with $A{A}_1=A_1{A}_2=A_2{A}_3$. Join $A_{3}B$. Through $A_1$ and $A_2$ draw lines parallel to $A_{3}B$, meeting $AB$ at $P$ and $Q$. Add the line through $A$ parallel to $A_{3}B$: four parallel lines make equal intercepts on $AX$, so they make equal intercepts on $AB$: $AP=PQ=QB=\frac{10}{3}$ cm.

11. Q11 · 4 marks · Long answerHOTS / competency

    In $△ABC$, $D$ is the mid-point of $AB$. Through $D$ draw $DE\parallel BC$ meeting $AC$ at $E$, and through $E$ draw $EF\parallel AB$ meeting $BC$ at $F$. Prove that $F$ is the mid-point of $BC$ and $DE=BF$.

    Hint

    Converse twice, then a parallelogram.

    Answer

    By the converse in $△ABC$, $E$ is the mid-point of $AC$. In $△CAB$, $E$ is the mid-point of $AC$ and $EF\parallel AB$, so $F$ is the mid-point of $BC$. $DEFB$ has $DE\parallel BF$ and $EF\parallel DB$, so it is a parallelogram and $DE=BF$.

12. Q12 · 3 marks · Short answerHarder exam level

    The diagonals of quadrilateral $ABCD$ are equal and perpendicular. What figure is formed by joining the mid-points of its sides? Give a reason.

    Hint

    Equal sides from equal diagonals; right angles from perpendicular diagonals.

    Answer

    A square: its sides are halves of equal diagonals (so all equal) and are parallel to perpendicular diagonals (so they meet at right angles).

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. 1.State the mid-point theorem.Show answer

   The segment joining the mid-points of two sides of a triangle is parallel to the third side and half of it.

2. 2.State the converse.Show answer

   A line through the mid-point of one side, parallel to another side, bisects the third side.

3. 3.State the equal intercept theorem.Show answer

   If a transversal makes equal intercepts on three or more parallel lines, any other transversal does too.

4. 4.Figure formed by joining mid-points of a quadrilateral?Show answer

   A parallelogram.

5. 5.Mid-line of a trapezium with parallel sides 8 and 14?Show answer

   11.

6. 6.Median to the hypotenuse of length 10?Show answer

   5.

7. 7.Perimeter of the mid-point triangle compared with the original?Show answer

   Half.

8. 8.What is an intercept?Show answer

   The part of a transversal between two of the lines it cuts.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Can you write the mid-point theorem proof with construction and reasons?
• ☐Can you prove the converse?
• ☐Can you prove the equal intercept theorem?
• ☐Do you know when to use the theorem and when the converse?
• ☐Can you prove that mid-points of a quadrilateral form a parallelogram?
• ☐Can you find a trapezium mid-line?
• ☐Can you prove AF = ⅓AC with the extra parallel line?

Mid-point and Intercept Theorems: frequently asked questions