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Reference: ML Agarwal – ICSE Class 9
Chapter 12: Rectilinear Figures
Rectilinear Figures
Syllabus at a Glance
ICSE Class 9 – Chapter 12: Rectilinear Figures
- Proof and use of theorems on parallelogram.
- Both pairs of opposite sides equal (without proof).
- Both pairs of opposite angles equal (with proof).
- One pair of opposite sides equal and parallel makes it a parallelogram (without proof).
- Diagonals bisect each other and bisect the parallelogram (with proof).
- Rhombus as a special parallelogram whose diagonals meet at right angles.
- In a rectangle, diagonals are equal; in a square they are equal and meet at right angles.
1. Revision of Basic Concepts: Quadrilaterals
What is a Quadrilateral?
[Diagram Placeholder]
A clean precise mathematical diagram. Background color: #e8f5e9 (light green). A general quadrilateral ABCD drawn on a #e8f5e9 background. Vertices labelled A, B, C, D in order. Sides AB, BC, CD, DA shown. Diagonals AC and BD drawn as dashed lines intersecting at point O. All four interior angles marked with different colored arcs. Thin dark lines, bold sans-serif font. Background strictly #e8f5e9.
A quadrilateral is a closed figure bounded by four line segments.
- It has 4 vertices, 4 sides, and 4 interior angles.
- It has 2 diagonals (line segments joining opposite vertices).
- Angle Sum Property: The sum of the four interior angles of a quadrilateral is always $360°$.
Types of Quadrilaterals
| Figure | Definition / Key Feature |
|---|
| Trapezium | A quadrilateral with exactly one pair of parallel opposite sides. |
| Parallelogram | A quadrilateral with both pairs of opposite sides parallel. |
| Rectangle | A parallelogram with one angle $90°$ (which makes all angles $90°$). |
| Rhombus | A parallelogram with all sides equal. |
| Square | A rectangle with all sides equal (or a rhombus with all angles $90°$). |
| Kite | A quadrilateral with two pairs of equal adjacent sides, but opposite sides are not equal. |
2. Theorems on Parallelogram
A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel. (Symbol: $||^{gm}$)
Let $ABCD$ be a parallelogram, where $AB \parallel DC$ and $AD \parallel BC$.
Theorem 1 — Diagonal bisects the Parallelogram
[Diagram Placeholder]
A precise mathematical diagram. Background color: #fff3e0 (light orange). A parallelogram ABCD on a #fff3e0 background. Diagonal AC is drawn, splitting it into two triangles, ABC and CDA. Parallel markings (arrows) on opposite sides: AB and DC have one arrow, AD and BC have two arrows. The two triangles are lightly shaded in different colors to show they are distinct but congruent. Label: "Diagonal AC divides parallelogram into two congruent triangles." Background strictly #fff3e0.
Statement: A diagonal of a parallelogram divides it into two congruent triangles.
Proof of Theorem 1
Given: A parallelogram $ABCD$ with diagonal $AC$.
To Prove: $\triangle ABC \cong \triangle CDA$.
Proof:
In $\triangle ABC$ and $\triangle CDA$:
$\angle BAC = \angle DCA$ (Alternate interior angles, since $AB \parallel DC$ and $AC$ is transversal) …(i)
$\angle BCA = \angle DAC$ (Alternate interior angles, since $AD \parallel BC$ and $AC$ is transversal) …(ii)
$AC = CA$ (Common side) …(iii)
Therefore, by ASA congruence rule, $\triangle ABC \cong \triangle CDA$. Hence Proved.
Theorem 2 — Opposite Sides are Equal
[Diagram Placeholder]
A precise mathematical diagram. Background color: #fff3e0 (light orange). A parallelogram ABCD on a #fff3e0 background. Opposite sides are marked as equal: AB and DC have a single red tick mark, AD and BC have double blue tick marks. Bold label: "Opposite sides are equal: AB = DC, AD = BC". Background strictly #fff3e0.
Statement: In a parallelogram, opposite sides are equal.
(Proof not required for ICSE, but it follows directly from Theorem 1 using CPCT).
Theorem 3 — Opposite Angles are Equal
[Diagram Placeholder]
A precise mathematical diagram. Background color: #fff3e0 (light orange). A parallelogram ABCD on a #fff3e0 background. Opposite angles are marked equal: Angle A and Angle C have a single blue arc, Angle B and Angle D have a double red arc. Bold label: "Opposite angles are equal: ∠A = ∠C, ∠B = ∠D". Background strictly #fff3e0.
Statement: In a parallelogram, opposite angles are equal.
Proof of Theorem 3
Given: A parallelogram $ABCD$.
To Prove: $\angle A = \angle C$ and $\angle B = \angle D$.
Proof:
Since $AB \parallel DC$ and $AD$ is transversal, consecutive interior angles are supplementary:
$\angle A + \angle D = 180°$ …(i)
Since $AD \parallel BC$ and $DC$ is transversal:
$\angle D + \angle C = 180°$ …(ii)
From (i) and (ii): $\angle A + \angle D = \angle D + \angle C \Rightarrow \angle A = \angle C$.
Similarly, we can prove $\angle B = \angle D$. Hence Proved.
Theorem 4 — Diagonals Bisect Each Other
[Diagram Placeholder]
A precise mathematical diagram. Background color: #fff3e0 (light orange). A parallelogram ABCD on a #fff3e0 background. Both diagonals AC and BD intersect at point O. Segments OA and OC have single red ticks (OA = OC). Segments OB and OD have double blue ticks (OB = OD). Bold label: "Diagonals bisect each other: OA = OC, OB = OD". Background strictly #fff3e0.
Statement: The diagonals of a parallelogram bisect each other.
Proof of Theorem 4
Given: A parallelogram $ABCD$ whose diagonals $AC$ and $BD$ intersect at $O$.
To Prove: $OA = OC$ and $OB = OD$.
Proof:
In $\triangle AOB$ and $\triangle COD$:
$\angle OAB = \angle OCD$ (Alternate interior angles, since $AB \parallel DC$)
$AB = DC$ (Opposite sides of parallelogram are equal)
$\angle OBA = \angle ODC$ (Alternate interior angles, since $AB \parallel DC$)
Therefore, by ASA congruence rule, $\triangle AOB \cong \triangle COD$.
By CPCT (Corresponding Parts of Congruent Triangles): $OA = OC$ and $OB = OD$. Hence Proved.
Theorem 5 — One pair of opposite sides equal and parallel
Statement: A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.
If $AB \parallel DC$ AND $AB = DC$, then $ABCD$ is a parallelogram.
(Proof not required for ICSE). This is extremely useful for proving a figure is a parallelogram quickly.
Worked Example 1 — Angles of a Parallelogram
Q. Two adjacent angles of a parallelogram are in the ratio 4 : 5. Find the measure of all its angles.
Adjacent angles in a parallelogram are supplementary.
Let angles be $4x$ and $5x$. Then $4x + 5x = 180° \Rightarrow 9x = 180° \Rightarrow x = 20°$.
Angles are $4(20°) = 80°$ and $5(20°) = 100°$.
Since opposite angles are equal, the four angles are $80°, 100°, 80°, 100°$.
Worked Example 2 — Diagonals Property
Q. In parallelogram $PQRS$, diagonals $PR$ and $QS$ intersect at $O$. If $PO = 3.5$ cm and $QO = 4.1$ cm, find the lengths of $PR$ and $QS$.
Diagonals bisect each other. Thus, $O$ is the midpoint of both $PR$ and $QS$.
$PR = 2 \times PO = 2 \times 3.5 = 7.0$ cm.
$QS = 2 \times QO = 2 \times 4.1 = 8.2$ cm.
Worked Example 3 — Proof using Theorem 5
Q. $ABCD$ is a parallelogram. $X$ and $Y$ are midpoints of opposite sides $AB$ and $DC$ respectively. Prove that $AXCY$ is a parallelogram.
Since $ABCD$ is a parallelogram, $AB \parallel DC$ and $AB = DC$.
Since $X$ and $Y$ are midpoints, $AX = \frac{1}{2}AB$ and $CY = \frac{1}{2}DC$.
Therefore, $AX = CY$. Also, $AX \parallel CY$ (parts of parallel lines).
In quadrilateral $AXCY$, one pair of opposite sides ($AX$ and $CY$) is equal and parallel.
Therefore, by Theorem 5, $AXCY$ is a parallelogram.
✏ Practice Problems — Parallelograms
- In a parallelogram $ABCD$, $\angle A = 65°$. Find the other three angles.
- The perimeter of a parallelogram is 150 cm. One side is greater than the other by 25 cm. Find the lengths of the sides of the parallelogram.
- In parallelogram $ABCD$, diagonals intersect at $O$. If $AC = 12.8$ cm and $BD = 7.6$ cm, find the measures of $OC$ and $OD$.
- $ABCD$ is a parallelogram. $E$ and $F$ are points on diagonal $BD$ such that $DP = BQ$. Show that $\triangle APD \cong \triangle CQB$ and hence prove $AP = CQ$.
- Two parallel lines $l$ and $m$ are intersected by a transversal $p$. Show that the quadrilateral formed by the bisectors of interior angles is a rectangle (which is a special parallelogram).
3. Special Parallelograms: Rhombus, Rectangle, Square
A parallelogram takes special names when it has additional specific properties.
Rhombus
[Diagram Placeholder]
A precise mathematical diagram. Background color: #e3f2fd (light blue). A rhombus ABCD on a #e3f2fd background. All four sides have a single tick mark to show they are equal. Diagonals AC and BD intersect at O. A small square at O indicates a right angle (90 degrees). Segments OA=OC (one tick on diagonals) and OB=OD (two ticks). Background strictly #e3f2fd.
A rhombus is a parallelogram in which all four sides are equal.
- It has all the properties of a parallelogram.
- Special Property: The diagonals of a rhombus bisect each other at right angles ($90°$).
- Special Property: The diagonals bisect the interior angles at the vertices.
Proof: Diagonals of a Rhombus meet at Right Angles
Given: A rhombus $ABCD$ with diagonals $AC$ and $BD$ intersecting at $O$.
To Prove: $\angle AOB = \angle BOC = \angle COD = \angle DOA = 90°$.
Proof:
Since a rhombus is a parallelogram, its diagonals bisect each other. Thus, $OA = OC$.
In $\triangle AOB$ and $\triangle COB$:
$OA = OC$ (proved above)
$AB = CB$ (sides of a rhombus are equal)
$OB = OB$ (common side)
By SSS congruence rule, $\triangle AOB \cong \triangle COB$.
By CPCT, $\angle AOB = \angle COB$.
Since $AC$ is a straight line, $\angle AOB + \angle COB = 180°$ (linear pair).
$2\angle AOB = 180° \Rightarrow \angle AOB = 90°$. Therefore, diagonals meet at right angles. Hence Proved.
Rectangle
[Diagram Placeholder]
A precise mathematical diagram. Background color: #e8f5e9 (light green). A rectangle ABCD on a #e8f5e9 background. All four interior angles have small squares indicating 90 degrees. Diagonals AC and BD are drawn. A bold label next to the diagonals: "AC = BD (Diagonals are equal)". Background strictly #e8f5e9.
A rectangle is a parallelogram in which one angle is $90°$ (which forces all four angles to be $90°$).
- It has all the properties of a parallelogram.
- All interior angles are $90°$.
- Special Property: The diagonals of a rectangle are equal in length.
Proof: Diagonals of a Rectangle are Equal
Given: A rectangle $ABCD$ with diagonals $AC$ and $BD$.
To Prove: $AC = BD$.
Proof:
In $\triangle ABC$ and $\triangle BAD$:
$AB = BA$ (common side)
$\angle ABC = \angle BAD = 90°$ (angles of a rectangle)
$BC = AD$ (opposite sides of a rectangle/parallelogram)
By SAS congruence rule, $\triangle ABC \cong \triangle BAD$.
By CPCT, $AC = BD$. Hence Proved.
Square
[Diagram Placeholder]
A precise mathematical diagram. Background color: #fff3e0 (light orange). A square ABCD on a #fff3e0 background. All four sides have single tick marks (equal). All four interior angles have 90-degree square symbols. Diagonals AC and BD intersect at O. A 90-degree square symbol is at the intersection O. Bold label: "Diagonals are EQUAL and bisect at RIGHT ANGLES." Background strictly #fff3e0.
A square is a rectangle with all sides equal, or a rhombus with all angles $90°$. It inherits all properties of a parallelogram, a rectangle, and a rhombus.
- All sides are equal. All angles are $90°$.
- Special Property: The diagonals of a square are equal AND they bisect each other at right angles.
Worked Example 4 — Rhombus Application
Q. The diagonals of a rhombus are 16 cm and 12 cm. Find the length of its side and its perimeter.
Diagonals of a rhombus bisect each other at right angles.
Let diagonals $AC = 16$ cm and $BD = 12$ cm meet at $O$. Then $OA = 8$ cm and $OB = 6$ cm, and $\angle AOB = 90°$.
In right $\triangle AOB$, by Pythagoras Theorem: $AB^2 = OA^2 + OB^2 = 8^2 + 6^2 = 64 + 36 = 100$.
$AB = 10$ cm. The side of the rhombus is 10 cm.
Perimeter $= 4 \times \text{side} = 4 \times 10 = 40$ cm.
Worked Example 5 — Rectangle Application
Q. In a rectangle $ABCD$, diagonal $AC$ is produced to $E$. $\angle BAC = 32°$. Find $\angle ACD$ and $\angle DCE$.
In rectangle $ABCD$, $AB \parallel DC$. $AC$ is the transversal.
Alternate interior angles are equal, so $\angle ACD = \angle BAC = 32°$.
$ACE$ is a straight line. Thus, $\angle ACD + \angle DCE = 180°$ (linear pair).
$32° + \angle DCE = 180° \Rightarrow \angle DCE = 148°$.
✏ Practice Problems — Special Parallelograms
- The diagonals of a rectangle $ABCD$ intersect at $O$. If $\angle BOC = 70°$, find $\angle ODA$.
- $ABCD$ is a rhombus with $\angle ABC = 56°$. Determine $\angle ACD$.
- Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.
- Show that the diagonals of a square are equal and bisect each other at right angles.
- In a square $PQRS$, diagonals intersect at $O$. Find the measure of $\angle POQ$ and $\angle OPQ$.
- The perimeter of a rhombus is 60 cm. If one of its diagonals is 18 cm long, find the length of the other diagonal.
4. Summary of Properties (Very Important for Exams)
Properties Checklist
| Property |
Parallelogram |
Rectangle |
Rhombus |
Square |
| Opposite sides parallel | ✓ Yes | ✓ Yes | ✓ Yes | ✓ Yes |
| Opposite sides equal | ✓ Yes | ✓ Yes | ✓ Yes | ✓ Yes |
| All sides equal | ✗ No | ✗ No | ✓ Yes | ✓ Yes |
| Opposite angles equal | ✓ Yes | ✓ Yes | ✓ Yes | ✓ Yes |
| All angles equal (90°) | ✗ No | ✓ Yes | ✗ No | ✓ Yes |
| Diagonals bisect each other | ✓ Yes | ✓ Yes | ✓ Yes | ✓ Yes |
| Diagonals are equal | ✗ No | ✓ Yes | ✗ No | ✓ Yes |
| Diagonals intersect at 90° | ✗ No | ✗ No | ✓ Yes | ✓ Yes |
| Diagonals bisect vertex angles | ✗ No | ✗ No | ✓ Yes | ✓ Yes |
Remember the hierarchy: A square is a rectangle. A square is a rhombus. A rectangle is a parallelogram. A rhombus is a parallelogram. But a parallelogram is not necessarily a rectangle or a rhombus!
