Plane Curve: Any drawing (straight or non-straight) done without lifting the pencil may be called a curve.

Closed Curves: Figures in which initial and endpoints coincide with each other are called closed figures.

Region: The interior of a curve together with its boundary is called its region.

Open Curves: A curve has two endpoints, and when it does not enclose the area within itself it is known as an open curve.

Simple Curves: Simple curve are those curves which changes direction but does not cross itself while changing direction.

Non-simple curve: If a curve does cross itself, then it is called a Non-simple curve.

A polygon is any closed, flat shape that is formed by straight line segments (sides).

Example:

• Trapezium: A trapezium is a quadrilateral where only two sides are parallel to each other.

Kite: A kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides that are adjacent to each other.

Parallelogram: A parallelogram is a quadrilateral whose opposite sides are parallel.

Rhombus: A rhombus is a quadrilateral with four equal-length sides and opposite sides parallel to each other.

Sum of interior angles of a polygon = (n – 2) × 180°

Explanation:

• A polygon with n sides can be divided into smaller triangles by drawing diagonals from one vertex.
• Each triangle has a total angle sum of 180°.
• Since the number of triangles formed inside the polygon is (n – 2).
• The total sum of all interior angles is therefore (n – 2) × 180°.

The opposite sides of a parallelogram are of equal length.

Given: ABCD is a parallelogram.

To Prove: AB = DC and BC = AD.

Construction: Draw any one diagonal, say `bar(AC)`.

Proof:

Consider a parallelogram ABCD,

In triangles ΔABC and ΔADC,

∠ 1 = ∠2, ∠ 3 = ∠ 4             .....(Pair of alternate angle)
and `bar(AC)` is common side.

Side AC = Side AC              .....(common side)
∠ 1 ≅ ∠2                             .....(Pair of alternate angle)
∠ 3 ≅ ∠ 4                            .....(Pair of alternate angle)

by ASA congruency condition,
∆ ABC ≅ ∆ CDA

This gives AB = DC and BC = AD.

Hence Proved.

The diagonals of a rhombus are perpendicular bisectors of one another.

Given: ABCD is a rhombus.

To Prove: m∠ AOD = m∠ COD = 90°.

Proof:

ABCD is a rhombus.                                                                         .........(Given)

Since the opposite sides of a rhombus have the same length, it is also a parallelogram.  ........(Properties of a rhombus)

The diagonals of a rhombus bisect each other.                            ..........(Properties of a rhombus)

Thus,                   

OA = OC            .....(DB ⟂ AC, Divides AO and OC into two equal parts)(1)

OB = OD            ......(AC ⟂ DB, Divides DO and OB into two equal parts)(2)

by SSS congruency criterion,
∆ AOD ≅ ∆ COD

Therefore, m∠ AOD = m ∠ COD.......(C.A.C.T.)

Since, ∠AOD and ∠ COD are a linear pair.

m∠ AOD = m∠ COD = 90°.

Hence Proved.

The diagonals of a rectangle are of equal length.

Given: ABCD is a rectangle. The diagonals are AC and BD bisect each other at a point O.

To prove: AC = BD 

Proof:
ABCD is a rectangle.
BC = AD                         ...........(Opposite sides are equal and parallel)(1)
m∠A = m∠ B = 90°.       ...........(Each of the angles is a right angle and opposite angles of a rectangle are equal.)(2)

Then looking at triangles ABC and ABD separately.

We have,

In ∆ ABC and ∆ABD,
AB = AB                       ......(Common side)
BC = AD                      ......(From 1)
m∠A = m∠ B = 90°.    ......(From 2)

by SAS congruency criterion,
∆ ABC ≅ ∆ ABD          .....(lies between two parallel lines)

Thus, AC = BD           ......(C.S.C.T.)

Hence Proved.

Given: A rectangle ABCD in which AR, BR, CP, DP are the bisects of ∠A, ∠B, ∠C, ∠D, respectively forming quadrilaterals PQRS.

To prove: PQRS is a square.

Proof:

In Δ ARB,

∠RAB + ∠RBA + ∠ARB = 180°

45° + 45° + ∠ARB = 180°

90° + ∠ARB = 180°

∠ARB = 180° - 90°

∴ ∠ARB = 90°

Similarly, ∠SRQ = 90°

In Δ ARB,

AR = BR  ...(i)

ΔASD ≅ Δ BQC   ...[By ASA rule]

AS = BQ  ...(ii)  [by CPCTC]

(i) - (ii)

AR - AS = BR - BQ

SR = RQ   ...(iii)

Also, SP = PQ  ...(iv)

PQ = RS  ...(v)

Hence, PQRS is a square.

The diagonals of a square are perpendicular bisectors of each other.

Given: ABCD is a square, where AC and BD is a diagonal bisect each other at a Point 'O'.

To Prove: ∠AOD = ∠COD = 90°.

Proof:

ABCD is a square whose diagonals meet at O. ......(Given)

OA = OC.                                                          ......(Since the square is a parallelogram)(1)

In ΔAOD and ∆COD,
OD = OD           .........(Common side)
OA = OC            .........(From 1)
AD = DC            ..........(All the sides of square have equal length.)

By SSS congruency condition,
∆AOD ≅ ∆COD

Therefore, m∠ AOD = m∠ COD  ......(C.A.C.T.)

Since, m∠ AOD and m∠ COD are a linear pair,

∠AOD = ∠COD = 90°.       

Hence Proved.