Wednesday, February 9, 2011

QUADRILATERALS 10 PROVES AND THEOREMS.


Theorem 8.1 : A diagonal of a parallelogram divides it into two congruent
triangles.
Proof : Let ABCD be a parallelogram and AC be a diagonal (see Fig. 8.8). 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. What do you observe?
You will find that AB = DC and AD = BC
Theorem 8.2 : In a parallelogram, opposite sides are equal.
You have already proved that a diagonal divides the parallelogram into two congruent
triangles; so what can you say about the corresponding parts say, the corresponding
sides? They are equal.
So, AB = DC and AD = BC
Now what is the converse of this result? You already know that whatever is given
in a theorem, the same is to be proved in the converse and whatever is proved in the
theorem it is given in the converse. Thus, Theorem 8.2 can be stated as given below :
If a quadrilateral is a parallelogram, then each pair of its opposite sides is equal. So
its converse is :
Theorem 8.3 : If each pair of opposite sides of a quadrilateral is equal, then it
is a parallelogram.
Can you reason out why?
Let sides AB and CD of the quadrilateral ABCD
be equal and also  AD = BC (see Fig. 8.9). Draw
diagonal AC.
Clearly, ∆ ABC ≅ ∆ CDA (Why?)
So, ∠ BAC = ∠ DCA
and ∠ BCA = ∠ DAC (Why?)
Can you now say that ABCD is a parallelogram? Why?
You have just seen that in a parallelogram each pair of opposite sides is equal and
conversely if each pair of opposite sides of a quadrilateral is equal, then it is a
parallelogram. Can we conclude the same result for the pairs of opposite angles?
Draw a parallelogram and measure its angles. What do you observe?
Each pair of opposite angles is equal.
Repeat this with some more parallelograms. We arrive at yet another result as
given below.
Theorem 8.4 : In a parallelogram, opposite angles are equal.
Now, is the converse of this result also true? Yes. Using the angle sum property of
a quadrilateral and the results of parallel lines intersected by a transversal, we can see
that the converse is also true. So, we have the following theorem :
Theorem 8.5 : If in a quadrilateral, each pair of opposite angles is equal, then
it is a parallelogram.
parallelogram ABCD and draw both its diagonals intersecting at the point O
(see Fig. 8.10).
Measure the lengths of OA, OB, OC and OD.
What do you observe? You will observe that
OA = OC and OB = OD.
or, O is the mid-point of both the diagonals.
Repeat this activity with some more parallelograms.
Each time you will find that O is the mid-point of both the diagonals.
So, we have the following theorem :
Theorem 8.6 : The diagonals of a parallelogram
bisect each other.
Now, what would happen, if in a quadrilateral
t h e   d i a g o n a l s   b i s e c t   e a c h   o t h e r ?  Wi l l   i t   b e   a
parallelogram? Indeed this is true.
This result is the converse of the result of
Theorem 8.6. It is given below:
Theorem 8.7 : If the diagonals of a quadrilateral
bisect each other, then it is a parallelogram.
You can reason out this result as follows:
Note that in Fig. 8.11, it is given that OA = OC
and OB = OD.
So,         ∆ AOB ≅ ∆ COD (Why?)
Therefore, ∠ ABO = ∠ CDO (Why?)
From this, we get AB || CD
Similarly,         BC || AD
Therefore ABCD is a parallelogram.
Theorem 8.8 : A quadrilateral is a parallelogram if a pair of opposite sides is
equal and parallel.
L o o k   a t   F i g   8 . 1 7   i n  wh i c h  AB   =   CD  a n d
AB || CD. Let us draw a diagonal AC. You can show
that ∆ ABC ≅ ∆ CDA by SAS congruence rule.
So, BC || AD (Why?)
Let us now take an example to apply this property of  parallelogram.
Theorem 8.9 : The line segment joining the mid-points of two sides of a triangle

is parallel to the third side.
You can prove this theorem using the following
clue:
Observe Fig 8.25 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 (Why?)
Therefore, BCDE is a parallelogram. (Why?)
This gives EF || BC.
In this case, also note that EF = 1/2ED = 1/2BC.
Can you state the converse of Theorem 8.9? Is the converse true?
You will see that converse of the above theorem is also true which is stated as
below:
Theorem 8.10 : The line drawn through the mid-point of one side of a triangle,
parallel to another side bisects the third side.In Fig 8.26, 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.


There is yet another property of a parallelogram. Let us study the same. Draw a

8 comments:

  1. thank you very much

    ReplyDelete
  2. This is the same as the text,
    But I need elaborated one

    ReplyDelete
  3. You Have just Note the statement that are written in the textbook I don't like it!!!

    ReplyDelete
  4. This is the same thing which is given in the text book . We want the correct proof .

    ReplyDelete