# Chapter – 6 : Properties Of Parallelogram | Chapter Solution Class 9

 Book Name : Ganit Prakash Subject : Mathematics (Maths) Class : 9 (Madhyamik/WB) Publisher : Prof. Nabanita Chatterjee Chapter Name : Properties Of Parallelogram (6th Chapter)

## Let us work out – 6.1

1. By calculating let us write the angles of the parallelogram ABCD when \mathbf{\angle B}=\mathbf{60^{\circ}}

Solution:

Given, \angle B=60^{\circ}\\

We have,

\angle B+\angle C=180^{\circ} \\

\text { or, } 60^{\circ}+\angle C=180^{\circ} \\

\text { or, } \angle C=180^{\circ}-60^{\circ} \\

\therefore \angle C=120^{\circ}\\

Again;

\angle C+\angle D=180^{\circ} \\

\text { or, } 120^{\circ}+\angle D=180^{\circ} \\

\text { or, } \angle D=180^{\circ}-120^{\circ} \\

\therefore \angle D=60^{\circ} \\

\therefore \angle B+\angle A=180^{\circ}\\

\text { or, }, 60^{\circ}+\angle A=180^{\circ}\\

\text { or, }, \angle A=180^{\circ}-60^{\circ}\\

\therefore \angle A=120^{\circ} \\

\therefore \angle A=120^{\circ},

\angle B=60^{\circ}, \angle C=120^{\circ},

\angle D=60^{\circ} (Ans)

2. In the picture of the parallelogram aside, let us calculate and write the value of \mathbf{\angle PRQ}\\

\therefore \angle SRQ+\angle P Q R=180^{\circ}\\

\text { or, } \angle S R P+\angle P R Q+\angle P Q R=180^{\circ} \\

\text { or, } 70^{\circ}+\angle P Q R+55^{\circ}=180^{\circ} \\

\text { or, } 125^{\circ}+\angle P R Q=180^{\circ}\\

\text { or, } \angle P R Q=180^{\circ}-125^{\circ}\\

\therefore \angle P Q R=55^{\circ}\\

3. In the picture aside, if AP and DP are the bisectors of \mathbf{\angle BAD} and \mathbf{\angle ADC} respectively of the parallelogram ABCD, then by calculating let us write the value of \mathbf{\angle APD}.

Solution:

We have,

\angle B A D=2 \angle P A D\\

and \angle C D A=2 \angle P D A\\

\therefore \angle B A D+\angle C D A=180^{\circ}\\

\text { or, } 2 \angle P A D+2 \angle P D A=180^{\circ}\\

\text { or, } 2(\angle P A D+\angle P D A)=180^{\circ}\\

\text { or, } \angle P A D+\angle P D A=\frac{180^{\circ}}{2}=90^{\circ}\\

In \triangle A P D\\

\therefore \angle APD =180^{\circ}-\angle P A D+\angle P D A\\

=180^{\circ}-90^{\circ} \\

=90^{\circ}(Ans)\\

4. By calculating, I write the values of X and Y in the following rectangles PQRS

Solution: (i)

Since diagonals of a rectangle are equal.

\therefore S Q=P R \\

\text { i.e, } S T=T Q \text { and } P T=T R \\

\therefore Q T=T R \\

\therefore \angle T Q R=\angle T R Q=25^{\circ} \\

X^{\circ}=\angle S R Q-\angle T R Q \\

\therefore P T=T S \\

\therefore \angle T P S=\angle P S T \\

\therefore \angle S Q R=\text { Alternate } \angle Q S P=25^{\circ} \\

\therefore \angle T P S=\angle P S T=25^{\circ} \\

\therefore Y^{\circ}=180^{\circ}-(\angle T P S+\angle P S T) \\

=180^{\circ}-\left(25^{\circ}+25^{\circ}\right) \\

=180^{\circ}-50^{\circ}=130^{\circ} \\

\therefore X^{\circ}=65^{\circ}, Y^{\circ}=130^{\circ}(\text { Ans })\\

(ii) Since diagonals of a rectangle are equal

\therefore S Q=P R \\

\text { i.e. } S T=T Q \text { and } P T=T R \\

\therefore Q T=T R \\

\angle T Q R=\angle T R Q \\

\angle T Q R+\angle T R Q+100^{\circ}=180^{\circ}\\

\text { or, } \angle T R Q+\angle T Q R=180^{\circ}-100^{\circ}\\

\text { or, } 2 \angle T Q R=80^{\circ}\\

\text { or, } \angle T Q R=\frac{80^{\circ}}{2}\\

\text { or, } \angle T Q R=40^{\circ}\\

\therefore Y^{\circ}=40\left[\because \angle T Q R=\text { Alternate } \angle Q S P=Y^{\circ}\right] \\

\therefore X^{\circ}+Y^{\circ}=90^{\circ} \\

\text { or, } X^{\circ}+Y^{\circ}=90^{\circ} \\

\text { or, } X^{\circ}+40^{\circ}=90^{\circ} \\

\text { or, } X^{\circ}=90^{\circ}-40^{\circ} \\

\therefore X^{\circ}=50^{\circ} \\

\therefore X^{\circ}=50^{\circ}, Y^{\circ}=40^{\circ}(\text { Ans })\\

5. In the figure aside, \mathbf{ABCD} are the two parallelograms. I prove with the reasons that \mathbf{CDFE} is also a parallelogram

Solution:

ABCD is a parallelogram \therefore A B \| C D\\

and A B=C D and A B E F is a parallelogram

A B \| E F and A B=E F \\

A B=C D and A B=E F\\

\therefore C D=E F \\

And A B \| C D and A B \| E F \\

\therefore C D \| E F\\

\therefore C D=E F and C D \| E F\\

\therefore C D E F is a parallelogram.(proved)\\

6. If in the parallelogram \mathbf{A B C D, A B>A D,} then I prove with reason that \mathbf{\angle B A C<\angle D A C}.\\

Solution:

Given: Let ABCD be a parallelogram such that A B>A D\\

\text { R.T:P: } \angle B A C<\angle D A \bar{C}\\

Proof: \quad \therefore A D \| B C \& A C is Diagonal

\therefore \angle D A C=\text { alternate } \angle A C B\\

In \triangle A B C,\\

or, \angle A B C>\angle A C B>\angle B A C\\

or, \angle A C B>\angle B A C\\

or, \angle D A C>\angle B A C\\

\therefore \angle B A C<\angle D A C (Proved)\\

1. Firoz had drawn a quadrilateral \mathbf{P Q R S} whose \mathbf{P Q=S R} and \mathbf{P Q R S}; I prove with the reason that \mathbf{PQRS} is a parallelogram

Solution:

P Q R S, P Q=S R \text { and } \\

P Q \| S R\\

R.T.P: PQRS is aparallelogram\\

Construction: Diagonal PR is drawn

Proof: In \triangle P Q R and \triangle P R S,\\

P Q=S R \text { Given }\\

\therefore \angle Q P R= alternate \angle P R S[\because P Q \| S R and P R is transversal]

PR is a common side

\therefore \triangle P Q R \cong \triangle P R S(B y S-A-S)\\

\therefore \angle P Q R=\angle R P S[ corresponding angles of acongruent triangles]

But as the transversal PR intersects the straight lines RQ and PS

and the two alternate angles become equal.

\therefore R Q \| P S \\

P Q \| S R \text { and } R Q \| P S\\

\therefore P Q R S is the parallelogram (proved).

2. Sabba has draw two straight lines such that, \mathbf{A D \| B C} and \mathbf{A D=B C}; I prove with reason that \mathbf{A B=D C} and \mathbf{A B} \| \mathbf{D C}

Solution:

A B C D, A D=B C \text { and } A D \| B C\\

R.T.P: A B=D C and A B \| D C\\

Construction:

Proof: In \triangle A B C and \triangle D A C\\

\angle A C B= alternate \angle D A C \cdot[\because \mathrm{AD} \| \mathrm{BC} and \mathrm{AC} is trasversal]\\

and AC is common side

\therefore \triangle \mathrm{ABC} \cong \triangle \mathrm{CDA} \text { (By } \mathrm{S}-\mathrm{A}-\mathrm{S}) \\

\therefore \angle \mathrm{BAC}=\angle \mathrm{ACD}

[Katex][\text { Corresponding angles of congurent } \\\text { traingles] }\\[/katex]

But as the transversal AC intersects the straight line AB and CD and the two alternate angles become equal

\therefore A B \| D C \text {. }\\

Since, in the quadrilateral, A B \| D C and B C \| A D\\

\therefore A B C D is a parallelogram

\therefore A B=C D\\

Hence A B=D C and A B \| D C (proved)\\

1. Let us prove that if the length of two diagonals of a parallelogram are equal, then the parallelogram will be rectangle.

Solution:

Given:

Let ABCD be a parallel log ram with two diagonals AC and BD such that AC = BD

R.T.P: ABCD is a rectangle

i.e. \angle A B C=90^{\circ}\\

Proof: \because A C=B D\\

\therefore A O=O B=O C=O D \\

\text { In } \triangle A O B \\

AO = OB

\therefore \angle O A B=\angle O B A \\

\text { In } \triangle O B C \\O B=O C \\

\therefore \angle O B C=\angle O C B\\

Again,\angle \mathrm{DAO}=\text { alternate }\\

\angle \mathrm{OCB}=\angle O B C \\

\therefore \angle \mathrm{OCB}=\angle O B C\\

We have

\angle D A B+\angle A B C=180^{\circ} \\

\text { or, } \angle D A O+\angle O A B+\angle O B A+\angle O B C=180^{\circ} \\

\therefore \angle O B C+\angle O B A+\angle O B A+\angle O B C=180^{\circ} \\

{[\because \angle D A O=\angle O B C \angle O A B=\angle O B A]} \\

\text { or, } 2 \angle O B C+2 \angle O B A=180^{\circ} \\

\text { or, } 2 \angle O B C+\angle O B A)=180^{\circ} \\

\text { or, } \angle O B C+\angle O B A=90^{\circ} \\

\text { or, } \angle A B C=90^{\circ}\\

Hence ABCD is a rectangle. (Proved)

2. Let us prove that, if parallelogram, the diagonals are equal length and intersects at right angles, the parallelogram will be square.

Solution:

Given:

Let ABCD be

a parallelogram with AC and BD are the two diagonals

\text{such that} A C=B D and \angle A O B=\angle B O C=\angle C O D=\angle D O A=90^{\circ}\\

R.T.P: ABCD is a square

\text { i.e. } \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{DA} \text { and }\angle \mathrm{ABC}=90^{\circ}\\

Proof: In the \triangle A O B and \triangle A O D,\\

\therefore \angle A O B=\angle D O A=90^{\circ}\\

\therefore OA is common

O D=O B[\because A C=B D \therefore O A=O B=O C=O D] \\

\therefore \triangle A O B \cong \triangle A O D[B y \mathrm{~A}-\mathrm{S}-\mathrm{S}]\\

Similarly, we can prove that,

AB = BC, BC = CD, CD = DA

Hence, AB = BC = CD = DA

Again, In \triangle B O C,\\

\angle O A B=\angle O B A=45^{\circ} \text { sum of three angles of a triangle } \\

\text { is } \left.180^{\circ}\right)\\

and In \triangle B O C,\\

\angle O B C=\angle O C B=45^{\circ} \\

\therefore \angle O B A+\angle O B C=45^{\circ}+45^{\circ} \\

\text { or, } \angle A B C=90^{\circ} \\

\therefore A B C D \text { is a square (proved) }\\

3. Let us prove that, a parallelogram whose diagonal intersect at right angle is a rhombus.

Solution:

Given:

Let ABCD be a parallegram with AC and BD are two diagonals such that AC = BD and \angle A O B=\angle B O C=\angle C O D=\angle D O A=90^{\circ}\\

RT.P: ABCD is a Rhombus.

i.e. AB = BC = CD = DA

Proof: In the \triangle A O B and \triangle A O D \\

\therefore \angle A O B=\angle D O A=90^{\circ}\\

\therefore OA is common OD=O B[\because A C=B D, \therefore O A=O B=O C=O D] \\

\therefore \triangle A O B \cong \triangle A O D[B y A-S-S] \\

\therefore A B=A D\\

Similarly, we can prove that, AB = BC, BC = C D \text { and } C D=D A\\

Hence, AB = BC = CD = DA

\therefore ABCD is a Rhombus. (Proved).

4. The two diagonal of a parallelogram intersect each other at the point O. A straight line passing through O intersects the sides AB and DC at the point P and Q respectively.Let us prove that \mathrm{OP}=\mathrm{OQ}

Solution:

Given:

Let ABCD be a parallelogram with AC and BD are the two diagonals such that AO = OC and OB = OD straight line passing through O intersects the sides AB and DC at the point P and Q respectively.

R.T.P: \mathrm{OP}=\mathrm{OQ}\\

Proof: In the \triangle A O P and \triangle C O Q\\

\therefore \angle A O P=\text { vertical } \angle C O Q \\

OC = OA

\therefore \angle O A P=\text { alternate } \\

\angle O C Q \\

\therefore \triangle A O P \cong \triangle C O Q \\

\therefore O P=O Q \text{operatorname({Proved }})\\

5. Let us prove that in an isosceles trapezium the two angles adjacent to any parallel sides are equal.

Solution:

Given:

Let ABCD be an isosceles

trapezium such that AB = CD \text { and } \mathrm{AD} \| \mathrm{BC} \\

R.T.P: \angle ABC = \angle BCD \text { and } \angle DAB = \angle ADC \\

Construction: we have to take a point Eon BC such that

E is the mid point of BC

i.e. \mathrm{BE}=\mathrm{EC}.\text{Join} \mathrm{A}, \mathrm{E} and \mathrm{E}, \mathrm{D}.\\

Proof: In \triangle ABE and \triangle CDE,\\

AB=CD \text { (Given) } \\

BE=EC(\text { Construction) } \\

AE=DE[\because A D \| B C] \\

\therefore \triangle A B C \cong \triangle C D E \\

\therefore \angle A B C=\angle BCD \text { and } \angle DAB=\angle ADC\\

i.e. the two angles adjacent to any parallel sides

are equal in an isosceles trapezium. (Proved)

6. In a square ABCD, P is any point on the side BC. The perpendicular drawn on AP from the point B intersects the sides DC at the point Q. Let us prove that AP = BQ.

Solution:

Given:

Let ABCD be a square,

\mathrm{P} is any point on the side BC. The perpendicular drawn on AP from the point B intersects

the side CD at the point \mathrm{Q}.

R.T.P: \mathrm{AP}=\mathrm{BQ}

Proof: In \triangle ABP and \triangle CBQ.

\angle CBA = \angle BCQ[\because ABCD] is square

AB = BC.

(Given)

\therefore \angle \mathrm{CQB}+\angle \mathrm{CPA}=180^{\circ} [\therefore \angle \mathrm{BCQ}=90^{\circ}, \mathrm{AP} \perp \mathrm{BQ}]......{(i})

\therefore \angle BPA+\angle C P A=180^{\circ}\\

[\because \text { BC is a straight line }]......{(ii)}\\

Subtract (ii) from (i) we have

\angle \mathrm{CQB}-\angle \mathrm{BPA}=0 \\

\therefore \angle \mathrm{CQB}=\angle \mathrm{BPA} \\

\therefore \triangle \mathrm{ABP} \cong \triangle C B Q[\therefore A-S-S] \\

\therefore A P=B Q \cdot \text {(proved)}.

7. Let us prove that, if the two opposite angles and two opposite sides of a quadrilateral are equal, then the quadrilateral will be a parallelogram.

Solution:

Given:

such that \angle ABC=\angle DCA and AB = CD\\

\therefore A B \| C D

R.T.P: ABCD is parallelogram.

Construction: Diagonal AC is drawn.

Proof: In \triangle ABC \text { and } \triangle CDA, AB = DC

\angle BAC = alternate \angle ACD [\because AB \| DC and A C is a transdversal]

and AC is common side.

\therefore \triangle ABC \cong \triangle CD A[B y S-A-S \text { Congruence }]

\angle ACB =\angle DAC [\text { corresponding angles of congruent triangles }]

But as the transversal AC inter sects the straight lines BC \text { and } A D

and the two alternate angles become equal

Since, the quadrilateral ABCD,AB \| DC \text { and } BC \| AD

\therefore ABCD is parallelogram.(Proved).

8. In the triangle \triangle A B C, the two medians B P and CQ are so extended upto the points R and S respectively that BP = PR and CQ = QS. Let us prove that, S, A, R are collinear.

Solution:

Given:

let ABC be a triangle, with two medians BP

and CQ are so extended upto the point R and S

respectively that BP = PR and CQ = QS

R.T.P: S, R, A, are Collinear

Construction: S, A; A, R; S, B and R, C are joined

proof: \therefore BP is median of \triangle ABC,

AP = PC

\text { and } B P=P R \text { (Given) }

\therefore A B C R \text { is a parallelogram }

\therefore B C \| A R \text {. }

Similarly, ACBS is a parallelogram

\therefore B C \| A S

\therefore B C \| A R \text { and } B C \| A S

\therefore B C \| S A R

This shows that S, A, R are collinear. (Proved)

9. The diagonal SQ of the parallelogram PQRS is divided i three equal parts at the point Kand L. PK intersects SQ at the point M and RL intersects PQ at the Point N. Let us prove that, PMRNis a parallelogram.

Solution :

Given:

Let PQRS be a parallelogram such that the diagonal SQ is divided into three equal parts at the parts at the points K and L . PK intersects SQ at

the point M and RL intersects PQ at the point N.

R.T.P: PMRN is a parallelogram

Proof: In the \triangle Q R L \text { and } \triangle P S K

QL = KS ( Given )

PS = QR (\therefore PQRS is a parallelogram)

\therefore \angle P S K=\text { alternate } \angle LQR

\therefore PS \| Q R \text { and } Q S \text { is transversal }] \\

\therefore \triangle QLR \cong \triangle P S K[B y S-S-A] \\

\therefore PK = RL

In the \triangle SKM and \triangle QLR\\

\therefore QL = KS(\text { Given })\\

\therefore \angle N Q L= alternate \angle \mathrm{MSK}[\therefore \mathrm{PQ} \| \mathrm{SR} \text { and } \mathrm{QS} is transversal]\\

\angle QNL =\angle S K M\\

\therefore \triangle SKM \cong \triangle QLN \\

\therefore NQ = SM

\therefore NQ = SM and PQ = PS

\therefore PN = MR

PN = MR \text { and } P N \| M R\\

\therefore PNMR is a parallelogram (proved)

10. In the parallelogram \mathrm{ABCD} \text { and } \mathrm{AECF}, \mathrm{AC} is a diagonal. If \mathrm{B}, \mathrm{E}, \mathrm{D}, \mathrm{F} are not collinear, then let us prove that, \mathrm{BEDF} is parallelogram.

Solution:

Given:

\text { Let } ABCD and AECF \text { be two parallelograms }

\text { Then, AD = BC, AB = CD },

A E = C F \text { and } A F=C E\\

R.T.P: B E D F  \text { is a parallelogram }\\

\text { i.e. } B F=E D \text { and } F D=B E\\

Proof: In \triangle A F D and \triangle B E C,\\

A D=B C \text { (Given) } \\

A F=C E(\text { Given ) } \\

\angle C A D=\text { alternate } \angle A C B \rightarrow(\text { i }) \\

\angle C A F=\text { alternate } \angle A C E \rightarrow(\text { ii })\\

(i) + (ii) Gives.....\\

\angle C A D+\angle C A F=\angle A C B+\angle A C E \\

\angle D A F=\angle B C E \\

\therefore \triangle A F D \cong \triangle B E C(B y S-A-S) \\

\therefore F D=B E\\

Again, In \triangle A D E and \triangle F C B ,\\

A D=B C \text { (Given }) \\

A E=F C(\text { Given })\\

\therefore \text { Remaining side } F B= \text { remaining side }  D E\\

\therefore F B=D E\\

Hence, B F=E D and F D=B E \\

\therefore B E D F \text { is a parallelogram } \\

11. ABCD  is quadrilateral The two parallelogram ABCE and BADF are drawn.Let us prove that, CD and EF bisect each other.

Solution:

Given:

Let ABCD be a quadrilateral and two parallelogram

R.T.P: CD and EF bisect each other at O

i.e. OE = OF and OC = OD.

Construction: C, E and E, F are join.

Proof: In the parallelogram ABCE ,

AB = EC \rightarrow(i)

In the parallelogram

ABFD , AB = DF \rightarrow(ii)

From (i) (ii) EC = DF

and AB \| EC and AB \| DF

\therefore EC \| DF

In \triangle OCE and \triangle ODF,

CE = DF

\triangle OEC=\text alternate \angle OFD

\angle EOC=\text vertical opposite \angle FOD

\therefore \triangle OCE \cong \triangle ODF

\therefore OE = OF \text and OC = OD

Hence CD and EF bisect each other(Proved).

12. In the parallelogram A B C D, A B=2 A D; Let us prove that, the bisectors of \angle B A D and \angle A B C meet at the mid point of the side DC in right angle.

Solution:

Given:

\therefore A B=2 A D \text { and } A B=C D, \\

\therefore B C=A D .\\

\text { i.e. } D P=\frac{1}{2} C D \text { and } \angle A P B=90^{\circ}\\

\therefore \angle D A P=\angle P A B=\text { alternate } \angle A P D \\

\therefore \therefore A B \| D C \text { and } A P \text { is transversal }] \\

\therefore A D=D P[\because \angle \mathrm{DAP}=\angle A P D] \\

\therefore D P=A D=\frac{1}{2} \mathrm{AB} \\

=\frac{1}{2} C D[\therefore A B=C D]\\

\therefore Pis the mid point of C D\\

Again,

\angle P A B+\angle P B A =\frac{1}{2}(\angle D A B+\angle C B A) \\

=\frac{1}{2} \times 180^{\circ}[\because A D \| B C] \\

=90^{\circ} \\

\therefore \angle A P B =180^{\circ}-(\angle P A B+\angle P B A) \\

=180^{\circ}-90^{\circ} \\

\therefore \angle A P B =90^{\circ}\\

13. The two squares ABPQ and ADRS drawn on two sides AB and AD of the parallelogram ABCD respectively are outside of ABCD. Let us prove that, PRC is an isosceles triangle.

Solution:

Given:

Let ABPQ and ADRS be two squares drawn two sides AB and AD of the parallelogram ABCD respctively are outside of ABCD.

R.T.P: PRC is an isosceles triangle

\text { i.e. } P C=R C

Proof: In the square ABPQ, AB = PB

In parallelogram ABCD, AB = CD

\therefore PB = CD

In parallelogram ABCD, A D=B C

\therefore BC= RD

\therefore PC =PB+B C

=C D+R D

=R C

\therefore P C =R C

\therefore P R C is an isosceles triangle(Proved)

14. In the parallelogram \mathrm{ABCD}, \angle \mathrm{BAD} is an obtuse angle; the two equilateral triangles \mathrm{ABP} and \mathrm{ADQ} are drawn on the two sides \mathrm{AB}\mathrm{AD} outside of it.Let us prove that, \mathrm{CPQ} is an equilateral triangle.

Solution:

Given:

Let ABCD be a parallelogram \angle B A D is an obtuse angle, the two equilateral triangles ABP and ADQ are drawn on the two sides AB & AD outside of it.

R.T.P: CPQ is an equilateral triangle.

Proof: Let us suppose, \angle A B C=x,

Then \angle B A D=180^{\circ}-x[\because A B C D is a parallelogram ]\\

\because A B P and A D Q are equilateral triangles

Then, \angle P A B=\angle A P B=\angle P B A=60^{\circ}\\

and \angle A Q D=\angle Q A D=\angle Q D A=60^{\circ}\\

In \triangle P B C and \triangle \mathrm{APQ},\\

\therefore P B=P A \text {. (Given }) \\

\therefore B C=A D=A Q(\text { Given }) \\

\angle P B C=\angle P A Q\left[\angle P B C=60^{\circ}+x\right. \\

\angle P A Q=360^{\circ}-\left(60^{\circ}+60^{\circ}+180^{\circ}-x\right)=60^{\circ}+x\\

\therefore P Q=P C\\

and \operatorname{In} \triangle P B C and \triangle C D Q ,\\

\therefore B C=A D=D Q(\text { Given }) \\

P B=A B=C D(\text { Given }) \\

\angle Q D C=60^{\circ}+x, \angle P B C=60^{\circ}+x \\

\therefore \angle Q D C=\angle P B C \\

\therefore P C=C Q\\

Hence P Q=P C= C Q\\

\thereforeC P Q is an equilateral triangle( proved).

15. OP, OQ and OR are three straight lines. The three parallelograms OPAQ, OQBR and ORCP are drawn. Let us prove that AR, BP,  and  CQ bisect each other.

Solution:

Given:

Let OPAQ, OQBR and ORCP be three parallelogram

Then, BQ = OR,

OR = CP

R.T.P: AR, BP and CQ bisect each other

Construction: P, B ; A, R and C, Q are joined and they meet at the point N.Join A, B and P, R.

Proof:\because B Q=O E \text { and } O R=C P \\

\therefore B Q=C P\\

and B Q \| O R and O R \| C P\\

\therefore B Q \| C P \text {. }\\

\because \angle C Q B= alternate \angle P C Q[\because B Q \| C P and C Q \text{is transversal}]

In \triangle B Q N and \triangle P N C\\

B Q=C P \\

\therefore \angle N Q B=\angle N C P \\

\therefore \angle B N Q=\text { vertical } \angle P N C \\

\therefore \triangle B Q N \cong \triangle P N C \\

\therefore B N=N P \text { and } Q N=N C\\

Smilarly, In \triangle A B N and \triangle P N R\\

\therefore A N=N R and B N=N P\\

Hence, A N=NR, B N=N P \ Q N=N C\\

\therefore A R, B P, and C Q \text{bisect each other (Proved)}

16. M.C.Q

(i) In the parallelogram \mathrm{ABCD}, \angle \mathrm{BAD}=75^{\circ} and \angle \mathrm{CBD}=60^{\circ}, then the value of \angle B D C is

(a) 60^{\circ}

(b) 75^{\circ}

(c) 45^{\circ}

(d) 50^{\circ}

Solution:

In the parallelogram,

\therefore \angle A=\angle C=75^{\circ} \\

\therefore \angle B D C =180^{\circ}-\left(75^{\circ}+60^{\circ}\right) \\

=180^{\circ}-135^{\circ} \\

=45^{\circ} \text { (c) is the correct answer. }

(ii) Let us write which of the following geometric figure has diagonals equal in length

(a) parallelogram

(b) Rhombus

(c) Trapezium

(d) Rectangle

(iii) In the parallelogram ABCD, \angle B A D=\angle A B C, the parallelogram is a

(a) Rhombus

(b) Trapezium

(c) Rectangle

(d) none of them

(iv) In the parallelogram ABCD, M is the mid point of the diagonal BD, if BM bisects \angle A B C, then the value of \angle A M B is

(a) 45^{\circ}

(b) 60^{\circ}

(c) 90^{\circ}

(d) 75^{\circ}

Solution:

\because M is the mid point of BD

\therefore B M=M D \text{and} \angle A B M=\angle M B C

\therefore \angle D B A=\angle M B C= \text{alternate} \angle B D C

In \triangle B D C

\therefore \angle D B C=\angle B D C

\therefore B C=D C

\therefore A B C D is a Rhombus

\therefore In Rhombus diagonal bisects at right angle each other

\therefore \angle A \cdot M B=90^{\circ}

(v) In the rhombus ABCD, \angle A C B=40^{\circ}, the value of \angle A D B is

(a) 50^{\circ}

(b) 110^{\circ}

(c) 90^{\circ}

(d) 120^{\circ}

Solution: \angle A C B=40^{\circ}

\angle A C B=\text { alternate } \angle C A D \\

\therefore \angle C A D=40^{\circ}, \angle D A O=40^{\circ}

In \triangle A O D,

\therefore \angle D A O =40^{\circ}, \angle A O D=90^{\circ} \\

\therefore \angle A D O =180^{\circ}-\left(40^{\circ}+90^{\circ}\right) \\

=180^{\circ}-130^{\circ}=50^{\circ} \\

\therefore \angle A D B =50^{\circ}

17. Short answer type Question :

(i) In the parallelogram \mathrm{ABCD}, \angle A: \angle B=3: 2. Let us write the measures of the angles of the parallelogram

Solution:

\text { Let } \angle A=3 x, \angle B=2 x \\

\therefore \angle A+\angle B=180^{\circ} \\

\text { or, } 3 x+2 x=180^{\circ} \\

\text { or, } 5 x=180^{\circ} \\

\text { or, } x=\frac{180^{\circ}}{5}=36^{\circ} \\

x=36^{\circ} \\

\therefore \angle A=3 x=3 \times 36^{\circ}=108^{\circ}=\angle C \\

\therefore \angle B=2 x=2 \times 36^{\circ}=72^{\circ}=\angle D

(ii) In the parallelogram ABCD, the bisectors of \angle A \text{and} \angle B \text{meet} \mathrm{CD} at the point \mathrm{E}. The length of the side BC is 2 \mathrm{~cm}. Let us write the length of the side AB.

Solution:

Since the bisectors of \angle \mathrm{A} \text{and} \angle B \text{meet} C D at the point E.So, A B= 2 B C.

\therefore A B =2 \times 2 \mathrm{~cm} \\

=4 \mathrm{~cm}

(iii) The equilateral triangle \mathrm{AOB} lie within the square ABCD. Let us write the value of \angle \mathrm{COD}.

\therefore \triangle A O B is an equilateral triangle

\therefore \angle A O B=\angle O A B=\angle O B A=60^{\circ} \\

\therefore \triangle D A B=90^{\circ} \quad \angle D A O=90^{\circ}-60^{\circ}=30^{\circ} \\

\therefore A O=A D[\because A O=A B]

Then, AOD is an isosceles triangle

\therefore \angle A O D=\angle A D O

We take \angle A O D=\angle A D O=x

\therefore \triangle A O D, \\

\text { or, } x+x+30^{\circ}=180^{\circ} \\

\text { or } 2 x=150^{\circ} \\

\therefore x=\frac{150^{\circ}}{2}=75^{\circ} \\

\because \triangle A O D=\triangle B O C \\

\therefore \angle A O D=\angle C O B=75^{\circ} \\

\therefore \angle C O D=360^{\circ}-(\angle A O D+\angle C O B+\angle A O B) \\

=360^{\circ}-\left(75^{\circ}+75^{\circ}+60^{\circ}\right) \\

=360^{\circ}-210^{\circ}=150^{\circ}(A n s)

(iv) In the square \mathrm{ABCD}, \mathrm{M} is a point on \mathrm{AD} so that \angle \mathrm{CMD}=30^{\circ} The diagonal BD intersects \mathrm{CM} at the point \mathrm{P}. Let us write the value of \triangle \mathrm{DPC}.

Solution:

Since ABCD is a square

\angle A=\angle B=\angle C=\angle D=90^{\circ} \\

\angle C M D =30^{\circ} \therefore \angle P D C=\frac{90^{\circ}}{2}=45^{\circ} \\

\therefore \angle D C M =180^{\circ}-\left(90^{\circ}+30^{\circ}\right) \\

=180^{\circ}-120^{\circ} \\

=60^{\circ} \\

\therefore \angle D P C =180^{\circ}-\left(60^{\circ}+45^{\circ}\right) \\

=180^{\circ}-105^{\circ} \\

=75^{\circ}(\text { Ans })

(v) In the Rhombus ABCD, the length of the side AB is 4cm \text{and} \angle B C D =60^{\circ}[/katex]. Let us write the length of the diagonal BD.

Solution:

Given, A B=4 \mathrm{~cm}, \angle B C D=60^{\circ}

\therefore \angle A B C+\angle B C D=180^{\circ} \\

\text { or, } \angle A B C+60^{\circ}=180^{\circ} \\

\therefore \angle A B C=180^{\circ}-60^{\circ}=120^{\circ}

Also, \angle D A B+\angle A B C=180^{\circ}

\text { or, } \angle D A B=180^{\circ}-\angle A B C \\

=180^{\circ}-120^{\circ}=60^{\circ} \\

\therefore \angle B C D=\angle D A B=60^{\circ} \\

\angle A B C=\angle A D C=120^{\circ} \\

Since diagonal BD bisects the angles \angle A B C \text{and} \angle A D C

\therefore \triangle A B D is an equilateral triangle

\therefore B D=4 \mathrm{~cm}(\text { Ans })

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