Quadrilaterals Class 9: Properties, Theorems & Mid-Point Theorem

A quadrilateral is a closed figure formed by four line segments. It has four sides, four vertices and four angles. The word comes from 'quad' (four) and 'lateral' (sides).

Examples you already know: squares, rectangles, parallelograms, rhombuses, trapeziums and kites. Chapter 8 of your NCERT textbook dives deep into the properties that connect these shapes.

Before we get to the theorems, let's nail the foundation — the angle sum property.

Theorem: The sum of the interior angles of a quadrilateral is $360^\circ$.

Proof:
Consider quadrilateral $ABCD$. Draw diagonal $AC$, which divides it into two triangles: $\triangle ABC$ and $\triangle ACD$.

By the angle sum property of a triangle:

Adding (i) and (ii):

Solved Example:
Three angles of a quadrilateral are $75^\circ$, $90^\circ$ and $110^\circ$. Find the fourth angle.

Understanding the hierarchy helps you remember which properties apply where.

Trapezium: A quadrilateral with one pair of parallel sides.

Parallelogram: A quadrilateral with both pairs of opposite sides parallel. ($AB \parallel DC$ and $AD \parallel BC$)

Rectangle: A parallelogram with all angles $= 90^\circ$.

Rhombus: A parallelogram with all sides equal.

Square: A parallelogram that is both a rectangle and a rhombus — all sides equal and all angles $= 90^\circ$.

Kite: A quadrilateral with two pairs of consecutive sides equal.

The hierarchy: Square $\subset$ Rectangle $\subset$ Parallelogram $\subset$ Quadrilateral, and Square $\subset$ Rhombus $\subset$ Parallelogram.

These are the core results of Chapter 8. Memorise these and their converses.

Property 1: Opposite sides of a parallelogram are equal.

Proof: In parallelogram $ABCD$, draw diagonal $AC$.

In $\triangle ABC$ and $\triangle CDA$:
- $\angle BAC = \angle DCA$ (Alternate interior angles, $AB \parallel DC$)
- $AC = CA$ (Common)
- $\angle BCA = \angle DAC$ (Alternate interior angles, $BC \parallel AD$)

$\therefore \triangle ABC \cong \triangle CDA$ (ASA)

By CPCT: $AB = CD$ and $BC = DA$ $\blacksquare$

Property 2: Opposite angles of a parallelogram are equal.
From the same congruence above, by CPCT: $\angle B = \angle D$. Similarly, $\angle A = \angle C$.

Property 3: Diagonals of a parallelogram bisect each other.

Proof: Let diagonals $AC$ and $BD$ intersect at $O$.

In $\triangle AOB$ and $\triangle COD$:
- $\angle OAB = \angle OCD$ (Alternate interior angles, $AB \parallel DC$)
- $AB = CD$ (Opposite sides of parallelogram)
- $\angle OBA = \angle ODC$ (Alternate interior angles)

$\therefore \triangle AOB \cong \triangle COD$ (ASA)

By CPCT: $OA = OC$ and $OB = OD$ $\blacksquare$

This means $O$ is the midpoint of both diagonals.

Rectangle: All properties of a parallelogram plus:
- All angles are $90^\circ$
- Diagonals are equal: $AC = BD$

Rhombus: All properties of a parallelogram plus:
- All sides are equal
- Diagonals bisect each other at right angles: $AC \perp BD$
- Diagonals bisect the vertex angles

Square: All properties of both rectangle and rhombus:
- All sides equal, all angles $90^\circ$
- Diagonals equal, bisect at right angles, and bisect vertex angles

Solved Example:
$ABCD$ is a rhombus. Show that diagonal $AC$ bisects $\angle A$ and $\angle C$.

Proof:
In $\triangle ABC$, $AB = BC$ (sides of rhombus).
$\therefore \angle BAC = \angle BCA$ (isosceles triangle property).

In $\triangle ADC$, $AD = DC$ (sides of rhombus).
$\therefore \angle DAC = \angle DCA$.

Since $\angle BAC = \angle BCA$ and $\angle DAC = \angle DCA$, diagonal $AC$ bisects both $\angle A$ and $\angle C$. $\blacksquare$

This is one of the most important and frequently tested theorems from Chapter 8.

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

Given: In $\triangle ABC$, $D$ and $E$ are the mid-points of $AB$ and $AC$ respectively.

To Prove: $DE \parallel BC$ and $DE = \frac{1}{2} BC$.

Proof:
Construct: Extend $DE$ to $F$ such that $DE = EF$. Join $CF$.

In $\triangle AED$ and $\triangle CEF$:
- $AE = CE$ ($E$ is midpoint of $AC$)
- $\angle AED = \angle CEF$ (Vertically Opposite Angles)
- $DE = EF$ (By construction)

$\therefore \triangle AED \cong \triangle CEF$ (SAS)

By CPCT:
- $AD = CF$ ...(i)
- $\angle ADE = \angle CFE$ ...(ii)

From (ii), $AD \parallel CF$ (alternate interior angles are equal).
Since $D$ is the midpoint of $AB$: $AD = DB$.
From (i): $CF = AD = DB$.

So $DB = CF$ and $DB \parallel CF$ (since $AD \parallel CF$ and $D$ lies on $AB$).

$\therefore DBCF$ is a parallelogram (one pair of opposite sides is equal and parallel).

Hence $DF \parallel BC$ and $DF = BC$.

Since $DE = \frac{1}{2} DF$:

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

Given: In $\triangle ABC$, $D$ is the mid-point of $AB$ and $DE \parallel BC$.

To Prove: $E$ is the mid-point of $AC$.

This converse is very useful in problems where you know a line is parallel to a side and passes through a midpoint — it immediately gives you another midpoint.

Solved Example:
$ABCD$ is a trapezium with $AB \parallel DC$. $E$ and $F$ are the mid-points of $AD$ and $BC$ respectively. Prove that $EF \parallel AB$ and $EF = \frac{1}{2}(AB + DC)$.

Solution:
Join $AC$. Let it intersect $EF$ at $G$.

In $\triangle ACD$: $E$ is the midpoint of $AD$ and $EG \parallel DC$ (since $EF \parallel AB \parallel DC$).
By the converse of the mid-point theorem, $G$ is the midpoint of $AC$ and $EG = \frac{1}{2} DC$.

In $\triangle ABC$: $G$ is the midpoint of $AC$ and $GF \parallel AB$.
By the mid-point theorem, $F$ is the midpoint of $BC$ (consistent with given) and $GF = \frac{1}{2} AB$.

Also, $EF \parallel AB$. $\blacksquare$

Example 1: $ABCD$ is a parallelogram. $AP$ and $CQ$ are perpendiculars from vertices $A$ and $C$ on diagonal $BD$. Prove that $AP = CQ$.

Proof:
In $\triangle APB$ and $\triangle CQD$:
- $\angle APB = \angle CQD = 90^\circ$ (Given)
- $AB = CD$ (Opposite sides of parallelogram)
- $\angle ABP = \angle CDQ$ (Alternate interior angles, $AB \parallel DC$)

$\therefore \triangle APB \cong \triangle CQD$ (AAS)

By CPCT: $AP = CQ$ $\blacksquare$

Example 2: Show that the line segments joining the mid-points of opposite sides of a quadrilateral bisect each other.

Proof:
Let $ABCD$ be a quadrilateral. Let $P, Q, R, S$ be the mid-points of $AB, BC, CD, DA$ respectively.

Join $AC$. In $\triangle ABC$, $P$ and $Q$ are mid-points of $AB$ and $BC$.
By mid-point theorem: $PQ \parallel AC$ and $PQ = \frac{1}{2} AC$ ...(i)

In $\triangle ACD$, $R$ and $S$ are mid-points of $CD$ and $DA$.
By mid-point theorem: $SR \parallel AC$ and $SR = \frac{1}{2} AC$ ...(ii)

From (i) and (ii): $PQ \parallel SR$ and $PQ = SR$.

$\therefore PQRS$ is a parallelogram.

Since diagonals of a parallelogram bisect each other, $PR$ and $QS$ bisect each other. $\blacksquare$

Ready to test your understanding? Try the SparkEd Quadrilaterals practice page for interactive problems that adapt to your level. Use the SparkEd Math Solver to verify your proofs or ask the SparkEd Coach for a hint when you are stuck.