Quadrilaterals -01

We already studied about the triangles in previous chapters. And we know that  joining 3 noncollinear points  in pairs we will get the shape triangle. Same way if we joints such a 4 points in an order we will get shape called quadrilaterals. Quadrilaterals have four side , four angle and four vertices. We can see the figures in page number 135.

Ina quadrilateral ABCD AB, BC, CD and DA are four sides. A,B,C,D are the 4 vertices. And ∠A,  ∠B,  ∠C,  ∠D, are the four angles formed at the vertices.

We can also say that the rectangle, square, parallelogram etc are the special type of quadrilaterals.

angle sum property of a quadrilateral

The sum of the angles of a quadrilaterals is 360°. This can be verify by drawing a diagonal AC in the quadrilateral ABC. This will be divide the quadrilateral in two triangles. fig 8.4 at p-136.

In triangle ADC ∠DAC +  ∠ACD +   ∠D = 180°

Similarly in triangle ABC ∠CAB +  ∠ACB +   ∠B = 180° Adding 1 and 2 we will get

∠DAC +  ∠ACD +   ∠D + ∠CAB +  ∠ACB +   ∠B = 180 +180 =360° Also we know

∠DAC + ∠CAB = ∠A,   ∠ACD +  ∠ACB = ∠C

so, ∠A +  ∠D + ∠B + ∠C = 360°.

type of quadrilaterals 


If one pair of the opposite side of a quadrilateral is parallel, then it is called trapezium.  ex. fig 8.5-1.


If both pair of opposite sides of a quadrilateral is parallel then it will called parallelograms. ex. Fig. 8.5 – 2,3,4,5.


If any one of the angle is 90° in a parallelogram, then it will be called rectangle. ex Fig 8.5 – 3.


If all sides of a parallelogram are same, then it will called as rhombus ex Fif 8.5 – 4.


If any one of the angle is 90° and all sides are same in a parallelogram, then it will call as square. ex. Fig 8.5 – 5.


If two pairs of adjacent sides of a quadrilateral is same, then it will call as kite. ex Fig.8.5 – 6.

Properties of parallelogram

Consider a parallelogram ABCD and diagonal AC. Cut the parallelogram through it diagonal AC. as shown in fig 8.7 in p-139, and we can see that it will divide the parallelogram in two triangle. And  place that two triangle one over other and can notice that both triangle are congruent.

 Theorem 8.1.

a diagonal of a parallelogram divide it into two congruent triangle.

Proof : Consider the parallelogram ABCD and AC is the diagonal of it. And the diagonal divide the parallelogram in two triangle ABC AND CDA. Now consider the BC parallel AD and AC transverse

so, ∠BCA=∠DAC (alternative interior angle)

Also AB parallel DC and AC transverse.

so, ∠BAC=∠DCA (alternate interior angle). And AC =CA (commenside)

There for triangle ABC congruent to triangle CDA (ASA rule).

Theorem 8.2.

In a parallelogram opposite side are equal.

this can be proved from above proof itself. Because the opposite side of a parallelogram are the corresponding sides / parts of a congruent triangle.

So, AB=DC and AD=BC.

Theorem 8.3.

If each pair of opposite side of a quadrilateral is equal, then it is a parallelogram.

This theorem is the converse of the theorem 8.2. Here also we can consider the above figure. From the figure assume that AB = CD and AD =BC. Now we can say the ΔABC and ΔCDA   are congruent (SSS).  and we can also observe that



Now ∠BAC + ∠DAC = ∠DCA + ∠BCA

Here the angle BAC + DAC and DCA + BCA  are the opposite corners of the quadrilateral . And we know that if the opposite angle are equal then the quadrilateral will be a parallelogram. and this one we can prove in the next theorem.

Theorem 8.4.

In a parallelogram , opposite angle are equal.

This can be prove by the same explanation of theorem 8.3. Consider AC is the diagonal of parallelogram ABCD. And by using the laws of parallel lines and transverse we can prove that



And hence opposite angle of parallelogram are equal.

Theorem 8.5.

This is the converse of theorem 8.4.

If a quadrilateral ,each pair of opposite angle are equal , then it is a parallelogram.

Theorem 8.6.

The diagonals of a parallelogram bisect each other.

This can be check by drawing diagonals of parallelogram and measure the intersecting point. and we can found that the diagonals will intersect at mid of the diagonal each other. This can be also prove by the theorems which we already studied. (parallel and transversal line and congruence of triangle.

Theorem 8.7.

This is the converse of the theorem 8.7

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

Proof :

From the fig.8.11 we can see that OA=OC and OB =OD so, Δ AOB congruent to ΔCOD (SAS, vertically opposite angle at O). There for ∠ABO and ∠CDO are same (CPCT). Then we can say that AB parallel to CD (parallel and transversal law). Same way  we can proof AD parallel to CB. And hence it is a parallelogram.

See page 141 for examples

Theorem 8.8

A quadrilateral is a parallelogram if a pair of opposite side is equal and parallel.

consider the figure 8.17 (p145) in which AB = CD and AB parallel to CD. and AC is the diagonal of the quadrilateral. ABC and CDA are congruent by SAS rule ( opposite interior angle). So BC  and AD  are equal and hence it is parallel. (equal line between parallel line will parallel). So this quadrilateral is parallel. see p-145 and 146 for examples and exercises.

The mid-point theorem

Draw a triangle ABC as shown in fig 8.24, mark the midpoint of the sides AB and AC with E and F respectively. Joint E and F and measure EF, BC, angle E and angle B. We can also repeat this experiment with different triangle also. And we can observe that EF = 1/2 BC, and angle AEF= angle ABC. And hence EF parallel to BC. (corresponding angles same).

Theorem 8.9

The line segment joining the mid-point of two side of a triangle is parallel to the third side.

Proof :    

In fig 8.25 E and F are midpoints of AB and AC respectively. And CD Parallel to BA . Δ AEF concourent to Δ CDF (ASA rule.) So EF=DF and BE=AE=DC. Also we know that EF = 1/2 BC.(midpoint theorem). there for each opposite pair are equal and hence BCDE is parallelogram and EF is parallel to BC

Theorem 8.10

This is converse of theorem 8.9, The line drawn through the mid-point of one side of triangle, parallel to another side bysects the third side.     See      p-149 and 150 for example and exercises.

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