MathsCBSE Class 9 · Quadrilaterals

The Midpoint Theorem and Its Converse: Class 9 Quadrilaterals

Updated 17 October 2026 · Fact-checked

The segment joining the midpoints of two sides of a triangle is parallel to the third side and half as long. Conversely, a line drawn through the midpoint of one side parallel to another side bisects the third side. Joining the midpoints of the sides of any quadrilateral in order gives a parallelogram (Varignon's parallelogram), whose sides are half the diagonals of the quadrilateral and parallel to them.

DE joining two midpoints is parallel to BC and half as long; joining a quadrilateral's midpoints gives a parallelogram with sides parallel to the diagonals.

Illustration drawn for QuestionBuddy.

Understand The Midpoint Theorem, Its Converse and Varignon's Parallelogram

Midpoint theorem. In $△ A B C$ let $D$ and $E$ be the midpoints of $A B$ and $A C$ . Then $D E ∥ B C$ and $D E = \frac{1}{2} B C$ .

*Proof sketch.* Extend $D E$ to $F$ so that $E F = D E$ and join $C F$ . In $△ A D E$ and $△ C F E$ : $A E = C E$ (midpoint), $∠ A E D = ∠ C E F$ (vertically opposite) and $D E = F E$ , so they are congruent (SAS). Then $A D = C F$ and $∠ A D E = ∠ C F E$ (alternate angles), so $C F ∥ A B$ , i.e. $C F ∥ B D$ . Since $A D = D B$ we get $C F = B D$ and $C F ∥ B D$ , so $B C F D$ is a parallelogram (one pair of opposite sides equal and parallel). Therefore $D F ∥ B C$ and $D F = B C$ ; as $D E = \frac{1}{2} D F$ , $D E = \frac{1}{2} B C$ and $D E ∥ B C$ .

Converse. If a line through the midpoint of one side of a triangle is parallel to another side, it bisects the third side. *Proof sketch.* Let $D$ be the midpoint of $A B$ , and the line through $D$ parallel to $B C$ meet $A C$ at $E$ . Draw $C F ∥ B A$ meeting $D E$ extended at $F$ . Then $D B C F$ is a parallelogram, so $C F = B D = A D$ . In $△ A D E$ and $△ C F E$ : $∠ A D E = ∠ C F E$ (alternate), $∠ A E D = ∠ C E F$ (vertically opposite), $A D = C F$ , so they are congruent (AAS) and $A E = C E$ .

Using the theorems. If $D E$ joins the midpoints of $A B$ and $A C$ and $D E = (2 x + 3)$ , $B C = (5 x - 2)$ , then $5 x - 2 = 2 (2 x + 3)$ gives $x = 8$ , $D E = 19$ and $B C = 38$ . The triangle with vertices at the three midpoints of the sides of a triangle with sides $14, 48, 50$ has sides $7, 24, 25$ , so its perimeter is $56$ , half of $112$ . For the triangle with sides $13, 14, 15$ (area $84$ ), the medial triangle has sides $6.5, 7, 7.5$ , perimeter $21$ and area $\frac{84}{4} = 21$ : the four small triangles formed by joining the midpoints are all congruent.

Varignon's parallelogram. In any quadrilateral $A B C D$ let $P, Q, R, S$ be the midpoints of $A B, B C, C D, D A$ . In $△ A B C$ , $P Q ∥ A C$ and $P Q = \frac{1}{2} A C$ ; in $△ A D C$ , $S R ∥ A C$ and $S R = \frac{1}{2} A C$ . So $P Q$ and $S R$ are equal and parallel, and $P QR S$ is a parallelogram. Its sides are half the diagonals, so with diagonals $12$ and $16$ the sides are $6$ and $8$ and the perimeter is $28$ . If the diagonals of $A B C D$ are equal, $P QR S$ is a rhombus; if they are perpendicular, $P QR S$ is a rectangle; if both, a square. In a rectangle with sides $12$ and $16$ the diagonals are both $20$ , so $P QR S$ is a rhombus of side $10$ .

Coordinates check. For $A (0, 0)$ , $B (6, 0)$ , $C (8, 6)$ , $D (2, 4)$ : the midpoints are $P (3, 0)$ , $Q (7, 3)$ , $R (5, 5)$ , $S (1, 2)$ and $P Q = S R = (4, 3)$ . The diagonals of $P QR S$ , $P R$ and $QS$ , share the midpoint $(4, 2.5)$ and $P Q = 5 = \frac{1}{2} A C$ where $A C = 10$ .

Method. Identify the midpoints, state the theorem used by name, convert the result to an equation (usually $segment = \frac{1}{2} \times side$ ), and for quadrilaterals draw a diagonal first.

The Midpoint Theorem, Its Converse and Varignon's Parallelogram: formulas to remember

Midpoint theorem: $D E ∥ B C, D E = \frac{1}{2} B C$; $D$ , $E$ midpoints of $A B$ , $A C$ .
Converse: $D E ∥ B C, D mid. of A B \Rightarrow E mid. of A C$; Line through a midpoint parallel to a side.
Medial triangle: $perimeter = \frac{1}{2} perimeter of △ A B C, area = \frac{1}{4}$; Joining the three midpoints.
Varignon's parallelogram: $P Q ∥ S R ∥ A C, P Q = S R = \frac{1}{2} A C$; Sides are half the diagonals.
Special Varignon figures: $A C = B D \Rightarrow rhombus; A C ⊥ B D \Rightarrow rectangle$; Both: square.
Trapezium from the midline: $area (D B C E) = \frac{1}{2} (D E + B C) h$; $D E = \frac{1}{2} B C$ .

How to solve The Midpoint Theorem, Its Converse and Varignon's Parallelogram questions

Look for midpoints, apply the theorem, and convert to an equation.

1Mark all midpoints given in the figure and join them if needed.
2Apply the midpoint theorem to the triangle containing the midpoints: write "parallel" and "half" separately.
3For a converse problem, check that the line is through a midpoint and parallel to a side, then conclude the other point is a midpoint.
4For a quadrilateral, draw a diagonal and apply the theorem in each of the two triangles.
5Compare the results to conclude a parallelogram, then add conditions (equal or perpendicular diagonals) for rhombus or rectangle.
6Check numbers: sides of the midpoint figure are half the corresponding sides or diagonals.

The Midpoint Theorem, Its Converse and Varignon's Parallelogram 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.

Midpoint theorem

Suppose a triangular garden has a base wall of 30 m, and a straight fence joins the midpoints of the other two sides.

How long is the fence and how is it placed relative to the base wall?

Try it first, then show the solutionSolution

By the midpoint theorem the fence is parallel to the base wall.
Its length is half of the base: $\frac{30}{2} = 15$ m.

Answer: The fence is 15 m long and parallel to the base wall.

Memory hook: Join the midpoints: half the length, parallel to the third side.

Tricks to remember The Midpoint Theorem, Its Converse and Varignon's Parallelogram

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

Parallel and half
Two midpoints joined: parallel to the third side and half as long, together.
Use it when: Midpoint theorem.
Midpoint plus parallel gives midpoint
From a midpoint, going parallel to another side lands on a midpoint.
Use it when: Converse.
Diagonals make the midpoint figure
Midpoint quadrilateral sides are half the diagonals, so the diagonals decide its shape.
Use it when: Varignon's parallelogram.

The Midpoint Theorem, Its Converse and Varignon's Parallelogram: solved examples

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

Example 1: Triangle of midpoints · 3 marks · Short answerBoard level

The sides of a triangle are 14 cm, 48 cm and 50 cm. Find the sides and the perimeter of the triangle formed by joining the midpoints of the sides, and decide whether it is right-angled.

Try it first, then show the solutionSolution

Each side of the midpoint triangle is half of the side of the original parallel to it. (1)
Sides: $7$ cm, $24$ cm and $25$ cm; perimeter $= 56$ cm, half of $14 + 48 + 50 = 112$ cm. (1)
$7^{2} + 2 4^{2} = 625 = 2 5^{2}$ , so it is right-angled (as is the original, since $1 4^{2} + 4 8^{2} = 2500 = 5 0^{2}$ ). (1)

Answer: The sides are 7 cm, 24 cm, 25 cm; the perimeter is 56 cm; it is right-angled.

Example 2: Algebra with the midpoint theorem · 3 marks · Short answerBoard level

In $△ A B C$ , $D$ and $E$ are the midpoints of $A B$ and $A C$ . $D E = (2 x + 3)$ cm, $B C = (5 x - 2)$ cm and $∠ A D E = 6 4^{\circ}$ . Find $x$ , $D E$ , $B C$ and $∠ A B C$ .

Try it first, then show the solutionSolution

$D E = \frac{1}{2} B C$ , so $5 x - 2 = 2 (2 x + 3) = 4 x + 6$ . (1)
$x = 8$ , $D E = 19$ cm and $B C = 38$ cm. (1)
$D E ∥ B C$ , so $∠ A B C = ∠ A D E = 6 4^{\circ}$ (corresponding angles). (1)

Answer: $x = 8$ , $D E = 19$ cm, $B C = 38$ cm, $∠ A B C = 6 4^{\circ}$ .

Example 3: Prove the converse · 5 marks · Long answerHarder board level

Prove that a line drawn through the midpoint of one side of a triangle, parallel to another side, bisects the third side.

Try it first, then show the solutionSolution

Given: $△ A B C$ , $D$ the midpoint of $A B$ , $D E ∥ B C$ with $E$ on $A C$ . To prove: $A E = E C$ . (½)
Construction: through $C$ draw $C F ∥ B A$ meeting $D E$ produced at $F$ . (½)
$D E ∥ B C$ and $C F ∥ B D$ , so $D B C F$ is a parallelogram; hence $C F = D B = A D$ . (1½)
In $△ A D E$ and $△ C F E$ : $∠ A D E = ∠ C F E$ (alternate angles, $A B ∥ C F$ ), $∠ A E D = ∠ C E F$ (vertically opposite), $A D = C F$ . (1½)
By AAS the triangles are congruent, so $A E = C E$ (CPCT): $E$ is the midpoint of $A C$ . (1)

Answer: $E$ is the midpoint of $A C$ .

Example 4: Varignon's parallelogram · 3 marks · Short answerHarder board level

The diagonals of a quadrilateral $A B C D$ are 12 cm and 16 cm. $P, Q, R, S$ are the midpoints of $A B, B C, C D, D A$ . Prove that $P QR S$ is a parallelogram, find its perimeter, and find its area if the diagonals of $A B C D$ are perpendicular.

Try it first, then show the solutionSolution

In $△ A B C$ , $P Q ∥ A C$ and $P Q = \frac{1}{2} A C$ ; in $△ A D C$ , $S R ∥ A C$ and $S R = \frac{1}{2} A C$ . So $P Q$ and $S R$ are equal and parallel: $P QR S$ is a parallelogram. (1)
Sides: $\frac{1}{2} (12) = 6$ cm and $\frac{1}{2} (16) = 8$ cm; perimeter $= 2 (6 + 8) = 28$ cm. (1)
If $A C ⊥ B D$ , then $P Q ⊥ QR$ , so $P QR S$ is a rectangle with area $6 \times 8 = 48$ cm $^{2}$ . (1)

Answer: $P QR S$ is a parallelogram of perimeter 28 cm; if the diagonals are perpendicular it is a rectangle of area 48 cm $^{2}$ .

Example 5: Dividing a triangular field · 4 marks · Case-basedHOTS / competency

Case: a triangular field $A B C$ has $B C = 80$ m and the height from $A$ to $B C$ is 60 m. A fence $D E$ joins the midpoints $D$ of $A B$ and $E$ of $A C$ . (i) How long is the fence and how is it positioned? (ii) What is the distance between $D E$ and $B C$ ? (iii) Find the area of the trapezium $D B C E$ . (iv) Find the area of $△ A D E$ and compare it with the area of $△ A B C$ .

Try it first, then show the solutionSolution

(i) $D E = \frac{1}{2} B C = 40$ m, parallel to $B C$ . (1)
(ii) $D E$ is halfway up, so its distance from $B C$ is $\frac{60}{2} = 30$ m. (½)
(iii) Area of $D B C E = \frac{1}{2} (40 + 80) (30) = 1800$ m $^{2}$ . (1)
(iv) $△ A D E$ has base 40 and height 30: area $= 600$ m $^{2}$ . $△ A B C$ has area $\frac{1}{2} \times 80 \times 60 = 2400$ m $^{2}$ ; $600 = \frac{1}{4} \times 2400$ . (1½)

Answer: (i) 40 m, parallel to $B C$ ; (ii) 30 m; (iii) 1800 m $^{2}$ ; (iv) 600 m $^{2}$ , one quarter of the field.

Where students lose marks in The Midpoint Theorem, Its Converse and Varignon's Parallelogram

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

Mistake: Stating only "half" or only "parallel"
The theorem has two parts.
Wrong working
$D E = \frac{1}{2} B C$ .
Right working
$D E ∥ B C$ and $D E = \frac{1}{2} B C$ .
Do this: Write both conclusions.
Usually costs: 1 mark.
Mistake: Using the converse without a midpoint
A line parallel to a side need not bisect another side.
Wrong working
$D E ∥ B C$ so $E$ is the midpoint of $A C$ .
Right working
$D$ must first be the midpoint of $A B$ .
Do this: State the midpoint condition.
Usually costs: The proof.
Mistake: Naming the wrong shape for Varignon's parallelogram
Not checking the diagonals.
Wrong working
The midpoint quadrilateral of a rectangle is a rectangle.
Right working
The diagonals of a rectangle are equal, so it is a rhombus.
Do this: Equal diagonals give a rhombus; perpendicular diagonals give a rectangle.
Usually costs: 2 marks.

Exam tips

Mark each midpoint with equal ticks on the figure.
Quote the theorem name each time: "midpoint theorem" or "converse of the midpoint theorem".
For a quadrilateral, draw one diagonal and use the theorem twice.
Check results with a numerical example when numbers are given.

Practice questions on The Midpoint Theorem, Its Converse and Varignon's Parallelogram

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

Q1 · 1 mark · MCQBoard level
In $△ A B C$ , $D$ and $E$ are the midpoints of $A B$ and $A C$ and $D E = 6.5$ cm. Then $B C$ equals
(a) 3.25 cm
(b) 13 cm
(c) 6.5 cm
(d) 19.5 cm
Hint
$D E$ is half of $B C$ .
Answer
(b) 13 cm. $B C = 2 \times 6.5 = 13$ cm.
Q2 · 3 marks · Short answerBoard level
$A B C D$ is a rectangle with $A B = 12$ cm and $B C = 16$ cm. $P, Q, R, S$ are the midpoints of its sides. Show that $P QR S$ is a rhombus and find its perimeter.
Hint
The diagonals of a rectangle are equal.
Answer
The diagonals are $A C = B D = 1 2^{2} + 1 6^{2} = 20$ cm. $P QR S$ is a parallelogram (Varignon) with sides $\frac{1}{2} A C = 10$ cm and $\frac{1}{2} B D = 10$ cm; all four sides are equal, so it is a rhombus. Perimeter $= 40$ cm.
Q3 · 3 marks · Short answerHarder board level
In $△ A B C$ , $D$ , $E$ , $F$ are the midpoints of $A B$ , $A C$ , $B C$ . Show that $D B F E$ is a parallelogram, and find its perimeter if $A B = 18$ cm and $B C = 24$ cm.
Hint
Use the midpoint theorem on $E F$ and $D E$ .
Answer
$E F ∥ A B$ (midpoints of $A C$ , $B C$ ) so $E F ∥ D B$ ; $D E ∥ B C$ so $D E ∥ B F$ . Both pairs of opposite sides are parallel: $D B F E$ is a parallelogram. $D B = 9$ cm, $B F = 12$ cm, so the perimeter is $2 (9 + 12) = 42$ cm.
Q4 · 5 marks · Long answerHarder board level
The vertices of a quadrilateral are $A (0, 0)$ , $B (6, 0)$ , $C (8, 6)$ and $D (2, 4)$ . Find the midpoints $P, Q, R, S$ of $A B, B C, C D, D A$ , show that $P QR S$ is a parallelogram, and verify that $P Q = \frac{1}{2} A C$ .
Hint
Use the midpoint formula and compare vectors $P Q$ and $S R$ .
Answer
$P = (3, 0)$ , $Q = (7, 3)$ , $R = (5, 5)$ , $S = (1, 2)$ . $P Q = (4, 3)$ and $S R = (4, 3)$ , so $P Q$ and $S R$ are equal and parallel: $P QR S$ is a parallelogram. The diagonals $P R$ and $QS$ both have midpoint $(4, 2.5)$ , confirming it. $P Q = 16 + 9 = 5$ and $A C = 64 + 36 = 10$ , so $P Q = \frac{1}{2} A C$ .
Q5 · 4 marks · Case-basedHOTS / competency
Case: a roof frame is a triangle $A B C$ with $A B = 13$ m, $B C = 14$ m and $C A = 15$ m. Beams join the midpoints $D, E, F$ of $A B, B C, C A$ . (i) Find the three beam lengths $D E$ , $E F$ , $F D$ . (ii) Find the total length of the beams. (iii) Find the area of $△ A B C$ by Heron's formula. (iv) Find the area of the beam triangle $D E F$ .
Hint
Each beam is half of the parallel side; the medial triangle has a quarter of the area.
Answer
(i) $D E = \frac{1}{2} C A = 7.5$ m, $E F = \frac{1}{2} A B = 6.5$ m, $F D = \frac{1}{2} B C = 7$ m. (ii) $7.5 + 6.5 + 7 = 21$ m. (iii) $s = 21$ , area $= 21 \times 8 \times 7 \times 6 = 84$ m $^{2}$ . (iv) The four small triangles are congruent, so $D E F$ has area $\frac{84}{4} = 21$ m $^{2}$ .

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.Midpoint theorem?Show answer
The segment joining the midpoints of two sides is parallel to the third side and half of it.
2.Converse?Show answer
A line through the midpoint of one side parallel to another side bisects the third side.
3.Perimeter of the medial triangle?Show answer
Half the perimeter of the original triangle.
4.Varignon's figure for a rectangle?Show answer
A rhombus.
5.Area of the medial triangle?Show answer
One quarter of the area of the original.

The Midpoint Theorem, Its Converse and Varignon's Parallelogram: frequently asked questions

Does the midpoint theorem work for any triangle?

Yes. It holds for every triangle, whatever its shape.

Why is the midpoint quadrilateral always a parallelogram?

Each pair of opposite sides is parallel to a diagonal of the original quadrilateral and half as long.

Is the converse of the midpoint theorem also called the intercept theorem?

The converse is a special case of the more general intercept theorem about parallel lines cutting transversals proportionally.