Quadrilaterals
Quadrilaterals
Quadrilaterals
CHAPTER 8
QUADRILATERALS
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.
There is yet another property of a parallelogram. Let us study the same. Draw a
parallelogram ABCD and draw both its diagonals intersecting at the point O
(see Fig. 8.4).
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 Fig. 8.4
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 the diagonals bisect each other?
Will it be aparallelogram? 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:
1 1 1
So, ∠ PAC + ∠ CAS = × 180° = 90°
2 2 2
or, ∠ BAC + ∠ CAD = 90°
or, ∠ BAD = 90°
So, ABCD is a parallelogram in which one angle is 90°.
Therefore, ABCD is a rectangle.
Example 5 : Show that the bisectors of angles of a parallelogram form a rectangle.
Solution : Let P, Q, R and S be the points of
intersection of the bisectors of ∠ A and ∠ B, ∠ B
and ∠ C, ∠ C and ∠ D, and ∠ D and ∠ A respectively
of parallelogram ABCD (see Fig. 8.10).
In ∆ ASD, what do you observe?
Since DS bisects ∠ D and AS bisects ∠ A, therefore, Fig. 8.10
1 1
∠ DAS + ∠ ADS = ∠ A+ ∠ D
2 2
1
= (∠ A + ∠ D)
2
1
= × 180° (∠ A and ∠ D are interior angles
2
on the same side of the transversal)
= 90°
Also, ∠ DAS + ∠ ADS + ∠ DSA = 180° (Angle sum property of a triangle)
or, 90° + ∠ DSA = 180°
or, ∠ DSA = 90°
So, ∠ PSR = 90° (Being vertically opposite to ∠ DSA)
Similarly, it can be shown that ∠ APB = 90° or ∠ SPQ = 90° (as it was shown for
∠ DSA). Similarly, ∠ PQR = 90° and ∠ SRQ = 90°.
So, PQRS is a quadrilateral in which all angles are right angles.
Can we conclude that it is a rectangle? Let us examine. We have shown that
∠ PSR = ∠ PQR = 90° and ∠ SPQ = ∠ SRQ = 90°. So both pairs of opposite angles
are equal.
Therefore, PQRS is a parallelogram in which one angle (in fact all angles) is 90° and
so, PQRS is a rectangle.
EXERCISE 8.1
1. If the diagonals of a parallelogram are equal, then show that it is a rectangle.
2. Show that the diagonals of a square are equal and bisect each other at right angles.
3. Diagonal AC of a parallelogram ABCD bisects
∠ A (see Fig. 8.11). Show that
(i) it bisects ∠ C also,
(ii) ABCD is a rhombus.
Fig. 8.13
Theorem 8.8 : 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 8.16 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. Fig. 8.16
1 1
In this case, also note that EF =ED = BC.
2 2
Can you state the converse of Theorem 8.8? Is the converse true?
You will see that converse of the above theorem is also true which is stated as
below:
Theorem 8.9 : The line drawn through the mid-point of one side of a triangle,
parallel to another side bisects the third side.
In Fig 8.17, 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.
Fig. 8.17
EXERCISE 8.2
1. ABCD is a quadrilateral in which P, Q, R and S are
mid-points of the sides AB, BC, CD and DA
(see Fig 8.20). AC is a diagonal. Show that :
1
(i) SR || AC and SR = AC
2
(ii) PQ = SR
(iii) PQRS is a parallelogram. Fig. 8.20
2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and
DA respectively. Show that the quadrilateral PQRS is a rectangle.
3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA
respectively. Show that the quadrilateral PQRS is a rhombus.
4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD.
A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.21). Show that
F is the mid-point of BC.
Fig. 8.21
5. In a parallelogram ABCD, E and F are the
mid-points of sides AB and CD respectively
(see Fig. 8.22). Show that the line segments AF
and EC trisect the diagonal BD.
Fig. 8.22
8.3 Summary
In this chapter, you have studied the following points :
1. A diagonal of a parallelogram divides it into two congruent triangles.
2. In a parallelogram,
(i) opposite sides are equal (ii) opposite angles are equal
(iii) diagonals bisect each other
3. Diagonals of a rectangle bisect each other and are equal and vice-versa.
4. Diagonals of a rhombus bisect each other at right angles and vice-versa.
5. Diagonals of a square bisect each other at right angles and are equal, and vice-versa.
6. The line-segment joining the mid-points of any two sides of a triangle is parallel to the
third side and is half of it.
7. A line through the mid-point of a side of a triangle parallel to another side bisects the third
side.