NCERT Solution Class 9th Maths Chapter – 8 Quadrilaterals Examples

Solution – Let us recall what a rectangle is. A rectangle is a parallelogram in which one angle is a right angle.
Let ABCD be a rectangle in which ∠ A = 90°.
We have to show that ∠ B = ∠ C = ∠ D = 90°
We have, AD || BC and AB is a transversal
(see Fig. 8.6).
So, ∠ A + ∠ B = 180° (Interior angles on the same
side of the transversal)
But, ∠ A = 90°
So, ∠ B = 180° – ∠ A = 180° – 90° = 90°
Now, ∠ C = ∠ A and ∠ D = ∠ B
(Opposite angles of the parallellogram)
So, ∠ C = 90° and ∠ D = 90°.
Therefore, each of the angles of a rectangle is a right angle

Solution – Consider the rhombus ABCD (see Fig. 8.7).
You know that AB = BC = CD = DA (Why?)
Now, in ∆ AOD and ∆ COD,
OA = OC (Diagonals of a parallelogram
bisect each other)
OD = OD (Common)
AD = CD
Therefore, ∆ AOD ≅ ∆ COD
(SSS congruence rule)
This gives, ∠ AOD = ∠ COD (CPCT)
But, ∠ AOD + ∠ COD = 180° (Linear pair)
So, 2∠ AOD = 180°
or, ∠ AOD = 90°
So, the diagonals of a rhombus are perpendicular to each other.

(i) ∠ DAC = ∠ BCA and (ii) ABCD is a parallelogram.

Solution – (i) ∆ ABC is isosceles in which AB = AC (Given)
So, ∠ ABC = ∠ ACB (Angles opposite to equal sides)
Also, ∠ PAC = ∠ ABC + ∠ ACB
(Exterior angle of a triangle)
or, ∠ PAC = 2∠ ACB (1)
Now, AD bisects ∠ PAC.
So, ∠ PAC = 2∠ DAC (2)
Therefore,
2∠ DAC = 2∠ ACB [From (1) and (2)]
or, ∠ DAC = ∠ ACB

(ii) Now, these equal angles form a pair of alternate angles when line segments BC and AD are intersected by a transversal AC.
So, BC || AD
Also, BA || CD (Given)
Now, both pairs of opposite sides of quadrilateral ABCD are parallel.
So, ABCD is a parallelogram

Solution – It is given that PS || QR and transversal p intersects them at points A and C respectively.
The bisectors of ∠ PAC and ∠ ACQ intersect at B and bisectors of ∠ ACR and
∠ SAC intersect at D.
We are to show that quadrilateral ABCD is a
rectangle.
Now, ∠ PAC = ∠ ACR
(Alternate angles as l || m and p is a transversal)

So, 1/2 ∠ PAC = 1/2 ∠ ACR
i.e., ∠ BAC = ∠ ACD
These form a pair of alternate angles for lines AB
and DC with AC as transversal and they are equal also.
So, AB || DC
Similarly, BC || AD (Considering ∠ ACB and ∠ CAD)

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

Solution – As D and E are mid-points of sides AB and BC of the triangle ABC, by Theorem 8.8,
DE || AC Similarly, DF || BC and EF || AB
Therefore ADEF, BDFE and DFCE are all parallelograms.
Now DE is a diagonal of the parallelogram BDFE,
therefore, ∆ BDE ≅ ∆ FED
Similarly ∆ DAF ≅ ∆ FED
and ∆ EFC ≅ ∆ FED
So, all the four triangles are congruent

Solution – We are given that AB = BC and have
to prove that DE = EF.
Let us join A to F intersecting m at G..
The trapezium ACFD is divided into two triangles;
namely ∆ ACF and ∆ AFD.
In ∆ ACF, it is given that B is the mid-point of AC (AB = BC)
and BG || CF (since m || n).
So, G is the mid-point of AF (by using Theorem 8.9)
Now, in ∆ AFD, we can apply the same argument as G is the mid-point of AF,
GE || AD and so by Theorem 8.9, E is the mid-point of DF,
i.e., DE = EF.
In other words, l, m and n cut off equal intercepts on q also