Revision: Geometry >> Quadrilaterals Maths English Medium Class 9 CBSE

We have a quadrilateral named PQRS, with diagonals PR and QS intersecting at O.

OP = 6.5 cm, OQ = 6.5 cm, OR = 5.2 cm, OS = 5.2 cm

We know that the diagonals of a parallelogram bisect each other.

But, here we have

OP ≠ OR

And OQ ≠ OS

Therefore,

PQRS is not a parallelogram.

Theorem: A quadrilateral is a parallelogram if a pair of opposite sides is equal and parallel.

In above fig. which is AB = CD and AB || CD .  Let us draw a diagonal AC. You can show that ∆ ABC ≅ ∆ CDA by SAS congruence rule. So ,  BC || AD . 

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.

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.

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°

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.

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: A diagonal of a parallelogram divides it into two congruent triangles.
Proof :  Let ABCD be a parallelogram and AC be a diagonal in following fig. 

Observe that the diagonal AC divides parallelogram ABCD into two triangles, namely, ∆ ABC and ∆ CDA. We need to prove that these triangles are congruent.
In ∆ ABC and ∆ CDA, note that BC || AD and AC is a transversal.
So, ∠ BCA = ∠ DAC (Pair of alternate angles)
Also, AB || DC and AC is a transversal.
So, ∠ BAC = ∠ DCA (Pair of alternate angles)
and AC = CA (Common)
So, ∆ ABC ≅ ∆ CDA (ASA rule)
or, diagonal AC divides parallelogram ABCD into two congruent triangles ABC and CDA. 
Now, measure the opposite sides of parallelogram ABCD. 
You will find that AB = DC and AD = BC.

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.