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.

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

• Isosceles trapezium: If the non-parallel sides of a trapezium are of equal length, we call it an isosceles trapezium.

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.

Theorem:  If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.
Proof: Let sides AB and CD of the quadrilateral ABCD be equal and also  AD = BC in following fig.

Draw diagonal AC. Clearly, ∆ ABC ≅ ∆ CDA 
So, ∠ BAC = ∠ DCA 
and ∠ BCA = ∠ DAC
Hence ABCD is a parallelogram.

Theorem:  If in a quadrilateral, each pair of opposite angles is equal, then it is a parallelogram.
Proof: In quadrilateral ABCD 
∠A = ∠D  
`=>` ∠A +∠B +∠C + ∠D
Now, ∠A +∠B +∠C + ∠D = 360°  (angle sum property of quadrilateral)
`=>` 2(∠A +∠B ) = 360°  
`=> ` ∠A +∠B  = 180°  
`therefore` ∠A +∠B  = ∠C + ∠D = 180°  
Line AB intersects AD and BC at A and B respectively. 
Such that ∠A +∠B = 180° 
`therefore`  AD || BC   (Sum of consecutive interior angle is 180° ) ...(1)
∠A +∠B  = 180°  
∠A +∠D  = 180°     (∠B= ∠D)
`therefore`  AB || CD  ..(2)
From (1) and (2), we get
AB || CD and AD || BC
`therefore` ABCD is a parallelogram.

Theorem : If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.
proof :  

In ∆ AOD and ∆ COB  (Given)
OD = OB   (Given) 
∠AOB = ∠COD (Vertically opposite angles are equal)
Therefore , ∆AOD ≅ ∆COB  (By SAS criterion of congruence)
So, ∠OAD = ∠OCB  ...(1) (C.P.C.T)
Now , lines AC intersects AD and BC at A and C respectively,
such that ∠OAD = ∠OCB  ...[from(1)]
`therefore` ∠OAD and ∠ OCB form a pair of alternate interior angles are equal.
Thus, AD || BC
Similarly , we can prove that AB || DC 
Hence , quadrilatera; ABCD is a parallelogram.

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.

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.

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 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.

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

Observe that E is the mid-point of AB, line l is passsing through E and is parallel to BC and CM || BA. 
Prove that AF = CF by using the congruence of ∆ AEF and ∆ CDF.

Let P and Q be the midpoints of AB and AC respectively.

Then PQ || BC such that

PQ = `1/2` BC         ......(i)

In ΔAPQ, D and E are the midpoint of AP and AQ are respectively

∴ DE || PQ and DE = `1/2` PQ       ....(ii)

From (1) and (2)   DE = `1/2 PQ = 1/2 PQ =  1/2  (1/2 BC) `    

DE = `1 /4`BC

Hence, proved.

We know that the diagonals of a rhombus are perpendicular bisectors of each other

∴ OA = OC

OB = OD

∠AOD = ∠COD = 90°

And ∠AOB = ∠COB = 90°

In ΔBDE, A and O are midpoints of BE and BD, respectively

OA || DE

OC || DG

In ΔCFA, B and O are midpoints of AF and AC, respectively

∴ OB || CF

OD || GC

Thus, in quadrilateral DOCG, we have

OC || DG and OD || GC

⇒ DOCG is a parallelogram

∠DGC = ∠DOC

∠DGC = 90°

Theorem : The line segment joining the mid-points of two sides of a triangle is parallel to the third side.
Observe following Fig.  in which E and F are mid-points of AB and AC respectively and CD || BA.

∆ AEF ≅ ∆ CDF (ASA Rule)
So, EF = DF and BE = AE = DC
Therefore, BCDE is a parallelogram.
This gives EF || BC. 
In this case, also note that EF = `1/2` ED = `1/2` BC.