Quadrilaterals: class 9 math: 9 Math

NCERT Exercise 8.1:Part-2: 9th math

Quadrilaterals Class 9 math NCERT Exercise 8.1 Question (5) Show that if the diagonals of a quadrilateral are equal and bisect each other at a right angle then it is a square.

Solution

Let PQRS is the given quadrilateral.

And PR and QS are its diagonal.

And as given, OP = OR and OS = OQ

And, ∠POS = ∠ROP = ∠ROQ = ∠ROS = 90^o

Then to prove PQRS is a square.

This means it is to be proved that

PQ=QR=RS=PS

And one of the internal angle = 90^o

In ΔPOS and ΔQOP

OS = OQ

[Because as given diagonals bisect each other]

∠POS = ∠QOP

[As per question]

And side OP is common in both of the triangle.

Therefore, by SAS congruency,

ΔPOS ≅ ΔQOP

Now, by CPCT we know that corresponding sides of congruent triangles are equal

Therefore, PS = PQ - - - - (i)

Now, in ΔQOP and ΔROQ,

OP = OR

[Because as given in question, diagonals bisect each other]

∠QOP = ∠ROQ = 90^o

[Because as per question diagonals bisect each other]

And side OQ is common in both of the triangles.

Therefore, by SAS (Side Angle Side) congruency,

ΔQOP ≅ ΔROQ

Now, by CPCT we know that corresponding sides of congruent triangles are equal

Thus, PQ = QR - - - - - (ii)

Similarly, in triangle ROQ and ΔROS

OQ = OS

[Because diagonals bisect each other as given in question]

And, ∠ROQ = ∠ROS = 90^o

[Because as per question diagonals bisect each other at right angle]

And side OR is common in both of the triangles.

Therefore, by SAS (Side Angle Side) congruency

ΔROQ ≅ Δ ROS

Now, by CPCT we know that corresponding sides of congruent triangles are equal

Therefore, QR = RS - - - - (iii)

Thus, by equation (i), (ii) and (iii) we get

PQ = QR = RS = PS (iv)

Now, in ΔPQS and ΔPRS

PQ = RS [from equation (iv)]

And PR = QS [As given in question, diagonals are equal]

And side PS is common in both of the triangles.

Therefore, by SSS (Side Side Side) congruency,

ΔPQS ≅ ΔPRS

Now, by CPCT we know that corresponding sides of congruent triangles are equal

Therefore, ∠RSP = ∠QPS - - - - - (v)

Now, PQ and SR are two lines and a transversal PS is intersecting it.

And we know that internal angles of the same side of a transversal are supplementary. This means the sum of angles of same side of a transversal is equal to 180^o

Therefore, ∠RSP + ∠QPS = 180^o

⇒ ∠RSP + ∠RSP = 180^o

[From equation (iv) ∠RSP = ∠QPS]

⇒ 2 ∠RSP = 180^o

⇒ ∠RSP = 180^o/2

⇒ ∠RSP = 90^o

Now, all sides of the given quadrilateral are equal and one of the internal angles is equal to 90^o, hence the given quadrilateral is a square Proved

Quadrilaterals Class 9 math NCERT Exercise 8.1 Question (6) Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that

(i) It bisects ∠C also.

(ii) ABCD is a Rhombus

Solution

Given, ABCD is a parallelogram.

And diagonal AC bisects ∠A

Therefore, ∠BAC = ∠CAD - - - - (i)

Thus, to prove

(i) It bisects ∠C also.

Here since ABCD is a parallelogram

Therefore, AB||DC and AD||BC

And, DC = AB and AD = BC

Now, since here a transversal AC intersects two parallel lilnes AB and DC

Therefore, alternate interior angles will be equal

i.e. ∠CAD = ∠BCA - - - - (ii)

And ∠BAC = ∠ACD - - - - (iii)

⇒ ∠CAD = ∠ACD - - - - (iv)

[From equation (i) ∠BAC = ∠CAD]

Now, from equation (ii) and equation (iv), we get

∠BCA = ∠ACD

Thus, given diagonal AC bisects ∠C also Proved

To prove (ii) ABCD is a parallelogram

In ΔACD

∠CAD = ∠ACD [From equation (iv)]

Now, we know that in a triangle sides opposide to equal angles are equal

Therefore, AD = CD - - - - (v)

Now, since ABCD is a parallelogram. And we know that opposide sides of a parallelogram are equal

i.e. AB = DC and AD = BC

Therefore, from equation (v)

AD = CD = BC = AB

Now, we know that a parallelogram in which all sides are equal is called a Rhombus.

Therefore given parallelogram ABCD is a Rhombus Proved

Quadrilaterals Class 9 math NCERT Exercise 8.1 Question (7) ABCD is a rhombus. Show that diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D.

Solution

Given, ABCD is a Rhombus

In which AC and BD are its diagonals.

Therefore, to prove that diagonal AC bisects ∠A and ∠C

And diagonal BD bisects ∠B and ∠D

This means to prove ∠1 = ∠2 and ∠3 = ∠4

And, ∠5 = ∠6 and ∠7 = ∠8

Now, In ΔABC,

AB = BC

[Because all sides of a Rhombus are equal and these are the sides of given Rhombus]

And we know that angles opposite to the equal sides are equal.

Therefore, ∠2 = ∠3 - - - - (i)

Now, AD||BC

[Because opposite sides of a Rhombus are parallel]

Here, AD||BC and a transversal AC is intersecting it, therefore alternate interior angles will be equal.

That is ∠2 = ∠4 - - - - (ii)

And, ∠1 = ∠3 - - - - (iii)

Now, from equation (i) and equation (ii), we get

∠3 = ∠4

And from equation (i) and equation (iii), we get

∠2 = ∠1

Thus, diagonal AC bisects ∠A as well as ∠C

Now, in ΔABD

Side AB = Side AD

[Since all sides of a Rhombus are equal]

And we know that angles opposite to the equal sides re equal

Therefore, ∠6 = ∠8 - - - - - (iv)

Now, since ABCD is a parallelogram

Thus, AB||CD

And a transversal BD is intersecting parallel lines AB and CD

Therefore alternate interior angles will be equal.

This means, ∠5 = ∠8 - - - - - (v)

And, ∠6 = ∠7 - - - - - - (vi)

Now, from equation (iv) and equation (v), we get

∠5 = ∠6

And, frome equation (iv) and equation (vi), we get

∠7 = ∠8

Thus, diagonal AC bisects ∠A as well as ∠C and diagonal BD bisects ∠B as well as ∠D Proved

Quadrilaterals Class 9 math NCERT Exercise 8.1 Question (8) ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that (i) ABCD is a square (ii) diagonal BD bisects ∠B as well as ∠D.

Solution

As given, Let ABCD is the given rectangle.

And AC and BD are its diagonals.

And AC bisects ∠A and ∠C

Then to prove

(i) ABCD is a square

Since, as given in question ABCD is a rectangle

Therefore, opposite angles are equal.

i.e. ∠A = ∠C

Therefore, `1/2/_A=1/2/_C`

⇒ ∠2 = ∠3

Now, we know that, in a triangle; sides opposite to equal angles are equal.

Thus, AB = BC - - - - (i)

Now, since as per question ABCD is a rectangle, and opposite sides of a rectangle are equal.

Therefore, AB = DC - - - - (ii)

And AD = BC - - - - - (iii)

Now, from equation (i), (ii) and (iii), we get

AB = BC = DC = AD

Here, since all sides of the given rectangle are equal, therefore ABCD is a square. Proved

To prove (ii) Diagonal BD bisects ∠B as well as ∠D

In ΔABD

AB = AD

[Since ABCD is a square as proved in section (i) and all sides of a square are equal]

Now, we know that, in a triangle angles opposite to the equal sides are equal

Therefore, ∠6 = ∠8 - - - - - (iv)

Here, since as given in question ABCD is a rectangle and we know that opposite sides of a rectangle are parallel.

Thus, here AB||DC and a transversal BD is intersecting it, therefore alternate interior angles will be equal.

i.e. ∠8 = ∠5 - - - - - (v)

And, ∠6 = ∠7 - - - - (vi)

Now, from equation (iv) and equation (v), we get

∠6 = ∠5

And from equation (iv) and equation (vi), we get

∠7 = ∠8

Thus, diagonal BD bisects ∠B as well as ∠D. Proved

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