# Chapter – 9 : Transversal and Mid Point Theorem | Chapter Solution Class 9

 Book Name : Ganit Prakash Subject : Mathematics (Maths) Class : 9 (Madhyamik/WB) Publisher : Prof. Nabanita Chatterjee Chapter Name : Transversal and Mid Point Theorem (9th Chapter)

## Let us work out – 9

1. In the triangle ABC, D is the midpoint of the side BC. From the point D, Paralle straight lines of CA and BA intersect at the point E and F respectively. Let us prove that, EF = \frac{1}{2} BC.

Solution:

Given:

Let ABC be a triangle and D is mid point of BC

DE \parallel CA and DF \parallel BA

R.T.P:  EF = \frac{1}{2} BC

Construction: Joint E and F

Proof: \therefore D is mid point of BC and DE \parallel CA

\therefore E is mid point of BA

\therefore D is mid point of BC and DF \parallel BA

\therefore F is the mid point of CA

Now, E and F are mid point of BA and CA

\therefore EF = \frac{1}{2} BC (proved)

2. D and E lie AB and AC respectively of the triangle ABC such that, AD = \frac{1}{4} and AE = \frac{1}{4} AC. Let us prove that, DE \parallel BC and DE = \frac{1}{4} BC.

Solution:

Given:

Let ABC be a triangle, D, and E lie on AB and AC respectively such that AD = \frac{1}{4} AB and AE = \frac{1}{4} AC

R.T.P: DE \parallel BC and DE = \frac{1}{4} BC

Construction: We take G & F are the mid point of AB and AC respectively and join G, F

Proof: \because F and G are the mid point of AB and AC of \bigtriangleup ABC

\therefore FG \parallel BC and GF = \frac{1}{2} BC

\therefore AD = \frac{1}{4} AB and AE = \frac{1}{4} AC

\therefore AG = GB and AF = FC

\therefore D and F are the mid point of Ag and AF respectively of \bigtriangleup  AFG

\therefore DE \parallel GF and GF \parallel BC

\therefore DE \parallel BC

and DE =  \frac{1}{4} GF

= \frac{1}{2} \times \frac{1}{2} BC

= \frac{1}{4}BC

\therefore DE = \frac{1}{4} BC (proved)

3. In the triangle PQR , the mid points of the sides QR and QP are X and Z respectively. The side QP is extended upto the point S so that PS = ZP. SX intersects the side PR at the point Y. Let us prove that, PY=\frac{1}{4} P R \\

Solution:

Given:

Let PQR be a triangle such that X and Z are mid points of QR and QP respectively. QP is extended upto the point S so that PS = ZP. SX intersects the side PR at the point Y.

R.T.P: P Y=\frac{1}{4} P R \\

Proof: \operatorname{In} \triangle P Q R, \\

X and Z are mid point of QR and QP respectively\therefore Z X \| P R \text { and } Z X=\frac{1}{2} P R \\

In \triangle S Z X, P Y \| Z X[\because Z X \| P R] \\

\therefore P Y=\frac{1}{2} Z X \\

=\frac{1}{2} \times \frac{1}{2} P R[\therefore Z X=\frac{1}{2} P R] \\

=\frac{1}{4} P R \\

\therefore P Y=\frac{1}{4} P R(\text { proved })\\

4. Let us prove that, the quadrilateral formed by joining midpoint of consecutive sides of a parallelogram.

Solution:

Given:

Let ABCD be a parallelogram such that K, L, M & N are the mid points of AB, BC, CD and AD respectively, KL, LM, MN and NK are drawn.

R.T.P: KLMN is a parallelogram

Construction: Join B & D

Proof: ln \triangle B D A \\

K and N are the mid points of A B \& A D \\

\therefore N K \| B D \text { and } N K=\frac{1}{2} B D\\

In\triangle BD C, L and M are the mid point of BC and CD

\therefore L M \| B D \text { and } L M=\frac{1}{2} B D \\

\therefore N K \| B D \text { and } L M \| B D \\

\therefore N K \| L M \\

\text { and } N K=\frac{1}{2} \mathrm{BD}, \mathrm{LM}=\frac{1}{2} \mathrm{BD} \\

\therefore N K=L M \\

\therefore N K \| L M \text { and } N K=L M\\

Hence KLMN is a parallelogram (proved).

5. Let us prove that, the quadrilateral formed by joining mid points of consecutive sides of a rectangular figure is not a square figure but a Rhombus.

Solution:

Given:

Let ABCD be a rectangle

Such that AB = CD, BC = AD.

K, L, M & N are the mid points of A B, B C, C D & A D respectively KL, LM MN are drown

R.T.P: KLMN is a Rhombus. But not a square.

Proof: ln \triangle A K N \ and \ \triangle B L K. \\

AK = BK[\therefore K is mid point of A B] \\

\mathrm{AN}=\mathrm{BL}[\therefore A B C D is a rectangle ] \\

\therefore \angle N A K=\angle K B L [Each being right angles] \\

\triangle A N K \cong \triangle B K L[\because B y S-A-S congruence ] \\

Again, In \triangle B K L and \triangle L C M, K L=L M \\

In \triangle L C M and \triangle M D N, L M=M N \\

In \triangle M D N and \triangle N A K, \mathrm{MN}=\mathrm{KN} \\

KL = KN = LM = MN

But K M>N L[B C \| K M \& B C=K M] \\

\therefore Diagonal of a square be equal

\therefore KLMN is a Rhombus (proved)

6. Let us prove that, the quadrilateral formed by joining midpoints of consecutive sides of a square.

Solution:

Given:

Let ABCD be a square P, Q, R \& S be the mid points of A B B C, C D, \& A D respectively. P Q, Q R, R S \& S P are drawn.

R.T.P. PQRS is a square.

Construction: P, R \& S, Q are joined

Proof: In \triangle A P S\ and \ \triangle P B Q,\\

\therefore A P=B P[\therefore P \text { is midpoint of } A B] \\

\therefore A S=B Q[\therefore A B C D \text { is Parallelogram } \\

\therefore A D=B C] \\

\angle S A P=\angle P B Q\left[\therefore \text { Each being } 90^{\circ}\right] \\

\therefore \triangle S A P \cong \triangle P B Q[B y S-A-S \text {congruence}]\\

In \triangle P B Q and \triangle C Q R \therefore P Q=Q R \\

In \triangle C Q R and \triangle R D S \therefore Q R=R S \\

In \triangle R D S and \triangle S A P \quad \therefore R S=P S \\

\therefore P S=P Q=Q R=R S\\

Also,A B=B C=C D=D A=S Q=P R \\

\therefore Hence PQRS is a square. (proved)

7. Let us prove that, the quadrilateral formed by joining mid points of a Rhombus is a rectangle.

Solution:

Given:

Let \mathrm{ABCD} be a Rhombus P, Q, and S be the mid points of A B, B C, C D and A D respectively P Q, Q R, R S, \& P S are drawn.

R.T.P.: PQRS is a rectangle

Construction: P, R and S, Q are joined.

Proof: In \triangle \mathrm{SAP} and \triangle \mathrm{QCR}.\\

\mathrm{AP}=\mathrm{CR} \\

\mathrm{AS}=\mathrm{QC} \\

\angle \mathrm{SAP}=\angle \mathrm{RCQ}[\therefore \text { Each right angle }] \\

\therefore \triangle \mathrm{SAP} \cong \triangle \mathrm{QCR} \text { [By } \mathrm{S}-\mathrm{A}-\mathrm{S} \text { congrunce] } \\

\therefore \mathrm{PS}=\mathrm{QR} \\

Similarly, In \triangle \mathrm{QPB} and \triangle \mathrm{SRD} \\

\therefore \mathrm{PQ}=\mathrm{SR}\\

Hence, \mathrm{PS}=\mathrm{QR} and \mathrm{PQ}=\mathrm{SR} \\

Also, \mathrm{PR}>\mathrm{SQ} \\

\therefore \mathrm{PQRS} is a rectangle ( proved)

8. In the triangle \mathrm{ABC}, the med points of \mathrm{AB} \ and \ \mathrm{AC} are D and E respectively the midpoints of CD and BD are P and Q respectively. Let us prove that, BE and PQ bisect each other.

Solution:

Given:

Let A B C be a triangle Dand Earethe midpoints of A B and A C respectively. The mid points of C D and B D are P and Q respectively B E and P Q are drown.

RT.P.: BE and PQ bisect each other

Construction: P E is drown, B, P, and C, Q are joined

Proof: \therefore D is mid point of AB

\therefore A D=B D=2 Q B[\text { Q is mid point of } B D, B D=2 Q B] \\

\because E and P are mid point of sides A C and D C of \triangle D A C \\

\therefore E P \| A D \text { and } E P=\frac{1}{2} A D \\

\text { i.e, } E P \| Q B \text { cmd } E P=\frac{1}{2} \cdot 2 O B \\

\therefore E P \| Q B \text { and } E P=Q B

\therefore P B Q E \text { is a parallelogram }

\therefore PBQE is a parallelogram

\therefore BE and PQ are diagonals of Parallelogram

\therefore BE and PQ bisects each other PQBE (proved).

9. In the triangle ABC, AD is perpendicular on the bisector of \angle A B C From the point D, a straight line DE parallel to the side BC is drawn which intersects the side \mathrm{AC} Cat the point \mathrm{E}. Let us prove that \mathrm{AE}=\mathrm{EC}.

Solution:

Given:

Let ABC be triangle, AD is perpendicular on the bisector of \angle A B C. From the point D, a straight line DE parallel to the side BC is drawn which intersects the side AC at the point E

R.T.P: A E=E C \\

Proof: In \triangle A B D \ and \ \triangle D B F. \\

\angle A B D=\angle D B F[ Bisector of \angle A B F] \\

BD is common

\angle ADB =\angle BDF[each being {90}\degree ]\\

By A-S-A, \triangle A B D \cong \triangle D B F \\

\therefore A D=D F\\

In \triangle A F C, D is mid point of A F \\

andD E \| F C \\

\therefore A E=E C(\text { proved })\\

10. In the triangle \mathrm{ABC}, \mathrm{AD} is median. From the point \mathrm{B} \ and \ \mathrm{C} two straight lines BR and CT, parallel to AD are drawn, which meet extended \mathrm{BA} \ and \ \mathrm{CA} at the point \mathrm{T} \ and \ \mathrm{R} respectively. Let us prove that \frac{1}{\mathrm{AD}}=\frac{1}{\mathrm{RB}}+\frac{1}{\mathrm{TC}}

Solution:

Given:

Let ABC be a triangle AD is median. From, the points B and C, Two straight lines BR and CT parallel to AD are drawn which meet extended BA and CA at the point T and R respectively.

R.T.P.: \frac{1}{A D}=\frac{1}{R B}+\frac{1}{T C} \\

Proof: In \triangle B C R, D is mid point of B C \ and \ A D \| B R \\

\therefore A D \| B R and A is the midpoint of CR

\therefore A D=\frac{1}{2} R B\\

In \triangle B C T , D is midpoint of B C and A D \| C T \\

\therefore A D \| C T and A is mid point of BT

\therefore A D=\frac{1}{2} C T \\

\therefore \frac{1}{2} R B=\frac{1}{2} C T \\

\therefore R B=C T \\

\therefore \frac{1}{A D}=\frac{1}{\frac{1}{2} R B}=\frac{2}{R B}=\frac{1}{R B}+\frac{1}{B R} \\

\therefore \frac{1}{A D}=\frac{1}{R B}+\frac{1}{T C}[\because T C=R B] \text { (proved) }\\

11. In the trapezium ABCD, A B \| DC \ and \ A B>D C; the midpoint of two diagonal AC and BD are E and F respectively. Let us prove that, E F=\frac{1}{2}(A B-D C )\\

Solution:

Given:

Let ABCD be a trapezium, AB \| D C and A B>D C. The midpoints of two diagonals AC and BE are E and F respectively. EF is drawn

R.T.P.: EF=(\frac{1}{2} A B-D C) \\

Construction: We produced F to G and E to H such that G is mid point such that G is mid point of BC and H is mid point of AD [ \therefore E and F are the mid points of AC and BD respectively]

Proof: \operatorname{In} \triangle A D B

H \ and \ F are the mid points of AD and BD respectively

\therefore H F=\frac{1}{2} A B -------(i)\\

In \triangle A B C, \\

G and E are the mid points of BC and AC respectively

E G=\frac{1}{2} A B----------(ii)\\

H F+E G=\frac{1}{2} A B+\frac{1}{2} A B\\

or, H E+E F+E F+F G=A B \\

or, 2 E F=A B-(H E+F G)---------(iii)\\

Now, In A C D, E \ and \ H are the mid points of AC and AD respectively

\therefore H E=\frac{1}{2} C D\\

In \triangle BCD, F and G are the mid points of BD and BC respectively.

\therefore F G=\frac{1}{2} C D \\

\therefore (iii) implies,

2 E F =A B-(\frac{1}{2} D C+\frac{1}{2} D C) \\

2 E F =A B-D C \\

\therefore E F =\frac{1}{2}(A B-D C)(\text { proved })\\

12. C is the mid point of the line segment AB, PQ is any straight line. The minimum distances of the line PQ from the point A, B and \mathrm{C} are \mathrm{AR}+\mathrm{BS}=2 \mathrm{CT}.\\

Solution:

Given:

Let AB be line segment C is the mid point of AB, PQ is any straight line. The minimum distances of the line P Q from the points A, B and C are AR, BS and CT respectively.

R.T.P.: AR + BS = 2 CT

Construction: AS is drawn which intersects CT at D.

Proof: \therefore AR, BS and CT are minimum distances of A, B and C from PQ

\therefore AR, BS and CT are parallel to each other.

In \triangle ABS, C is mid point of AB and CD \| B S \\

\therefore D is mid point of AS  and \ CD=\frac{1}{2} B S i.e; B S=2 C D \\

Now, In \triangle S A R, D is mid point of A S \ and \ D T \| A R \\

\therefore D T =\frac{1}{2} A R \\

\text { i.e } ; A R =2 D T \\

\therefore A R+B S =2 D T+2 C D \\

=2(D T+C D) \\

=2 C T \quad[\therefore C T=D T+C D] \\

\therefore A R+B S =2 C T \quad(\text { proved }).\\

13. In a triangle \mathrm{ABC}, D is the midpoint of the side \mathrm{BC}, through the point \mathrm{A}, \mathrm{PQ} is any straight line The perpendicular from the point B, C and D on \mathrm{PQ} are BL, QM and DN respectively, let us prove that, DLDM

Solution:

Given:

Let ABC be a triangle, D is the mid points of BC, through the point A, PQ is any straight line, The perpendicular from the points B, C and D on PQ are BL, CM and DN respectively DL and DM are drawn

R.T.P.: DL = DM

Proof: In \triangle LND and \triangle N M D \triangle N D=\angle M N D[\text { each being } 90^{\circ}] \\

\because BD=DC[\because \text { D\ is mid point of } B C \\

\therefore LN=NM[B L \perp L M, N D \perp L M \ M C \perp L M] \\

\therefore ND is common

\therefore \triangle N D \cong \triangle N M D[\text { By } S-A-\text { S \ congrience }] \\

\therefore DL=D M(\text { proved }) \text {. }\\

14. ABCD is a squared figure. The two diagonals AC and BD intersects each other at the point \mathrm{O}. The bisector of \angle \mathrm{BAC} intersect \mathrm{BO} at the point \mathrm{P} \ and \  \mathrm{BC} at the point Q. Let us prove that, \mathrm{OP}=\frac{1}{2} \mathrm{CQ} \\

Solution:

Given:

Let ABCD be a Square, the two diagonals AC and BD intersects each other at the point O. The bisector of \angle B A C intersects BO at the point P and BC at the point Q.

Construction: We extend the line AQ to E and joined C & E. Such that CQ = CE and P is mid point AE.

Proof: In \triangle CQE, CQ = CE \text { (By construction) } \\

\angle CQE =\angle CEQ (\text { By construction) })\\

\angle C Q E=\text { Vertical opposite } \angle A Q B \\

\therefore \angle C A Q=\angle Q A B \\

\because \text { O is mid point of } A C \\

\text { and } P \text { is mid point of } A E \\

\therefore O P \| C Eand P is mid point of AE.

\therefore O P \| C E\\

By mid point Theorem,

O P =\frac{1}{2} C E \\

\therefore O P =\frac{1}{2} C Q[\quad C E=C Q] \quad \text { (proved) }\\

## 15. M.C.Q.

(i) In the triangle \mathrm{PQR}, \angle \mathrm{PQR}=90^{\circ} and \mathrm{PR}=10 \mathrm{~cm} . If S is the mid point of PR, then the length of QS is

(a) 4 \mathrm{~cm} \\

(b) 5 \mathrm{~cm} \\

(c) 6 \mathrm{~cm} \\

(d) 3 \mathrm{~cm} \\

Solution:

Given: \angle P Q R=90^{\circ} \ and \ P R=10 \mathrm{~cm} \\

S is the mid point of PR

\therefore P S =S R=\frac{10}{2} \mathrm{~cm}=5 \mathrm{~cm} \\

\therefore Q S =\frac{1}{2} P R \\

=\frac{1}{2} \times 10 \mathrm{~cm}=5 \mathrm{~cm} \\

\therefore(b) is correct option.

(ii)In the trapezium \mathrm{ABCD}, \mathrm{AB} \| \mathrm{DC} \ and \ \mathrm{AB}=7 \mathrm{~cm} and \mathrm{DC}=5 \mathrm{~cm}. The mid points of AD and BC are E and F respectively, the length of EF

(a) 5 \mathrm{~cm} \\

(b) 6 \mathrm{~cm}\\

(c) 7 \mathrm{~cm} \\

(d) 12 \mathrm{~cm} \\

Solution:

Given: A B=7 \mathrm{~cm}, D C=5 \mathrm{~cm}

We have,

E F =\frac{1}{2}(A B+D C) \\

=\frac{1}{2}(7+5) \mathrm{cm} \\

=\frac{1}{2} \times 12=6 \mathrm{~cm} \\

\therefore( b) is correct option.

(iii) In the triangle \mathrm{ABC}, \mathrm{E} is the midpoint of the median AD; the extended B E intersects AC at the point F. If AC = 10.5 cm then the length of AF is

(a) 3 \mathrm{~cm} \\

(b) 3.5 \mathrm{~cm} \\

(c) 2.5 \mathrm{~cm} \\

(d) 5 \mathrm{~cm} \\

Solution:

We have,

A F =\frac{1}{3} A C \\

=\frac{1}{3} \times 10.5 \\

=3.5 \mathrm{~cm} \\

\therefore(b) is correct option.

(iv) In the triangle ABC, the midpoint of BC, CA and AB are D, E and F respetively BE and DF intersects at the point X and C F and DE intersects at the point. Y, the length of XY is equal to

(a) \frac{1}{2} \mathrm{BC} \\

(b) \frac{1}{4} \mathrm{BC} \\

(c) \frac{1}{3} \mathrm{BC} \\

(d) \frac{1}{8} BC \\

Solution:

Since D, E & F are the mid points of BC, CA & A B respectively\therefore E F \| B C \\

E F=\frac{1}{2} B C \\

\therefore X Y \| E F \\

\therefore X Y=\frac{1}{2} E F \\

=\frac{1}{2} \times \frac{1}{2} B C=\frac{1}{4} B C \\

\therefore(b) is correct option.

(v) In the parallelogram ABCD, E is the mid point of the side BC ; DE and extended A B meet at the point F. The length of AF is equal to

(a) \frac{3}{2} \mathrm{AB}

(b) 2 \mathrm{AB}

(c) 3 \mathrm{AB}

(d) \frac{5}{4} \mathrm{AB}

Solution:

In \triangle B E F \ and \ \triangle C E D

\angle C E D= vertically opposite \angle E B F

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

\therefore E C=B E (Given)

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

\therefore C D=B F=A B[A B=C D]

\therefore A F=A B+B F =A B+A B =2 A B

\therefore(b) is correct option.

(i) In the triangle ABC, AD and BE are two medians and DF parallel to BE, meets AC at the point F. If the length of the side AC is 8 cm. Then let us write the length of the side CF.

Solution:

A E=E C \\

B E \| D F \\

\therefore E F=F C \\

\therefore C F=\frac{1}{2} E C \\

=\frac{1}{2} \times \frac{1}{2} \times A C \\

=\frac{1}{4} \times 8 \mathrm{~cm}=2 \mathrm{~cm} \cdot(\text { Ans })

(ii) In the triangle  ABC, the mid point of BC, CA and AB are P, Q and R respectively if AC = 21 cm, BC = 29 cm, and AB = 30 cm, then let us write the perimeter of the quadrilateral ARPQ.

Solution:

A C=21 \mathrm{~cm}, B C=29 \mathrm{~cm} \\

A B=30 \mathrm{~cm} \\

\therefore A Q=\frac{A C}{2}=\frac{21}{2} \mathrm{~cm} \\

A R=\frac{A B}{2}=\frac{30}{2} \mathrm{~cm}=15 \mathrm{~cm} \\

\because \mathrm{QP} \| \mathrm{AB} \\

\mathrm{QP}=\frac{1}{2} A B=\frac{1}{2} \times 30 \mathrm{~cm}=15 \mathrm{~cm} \\

\because P R \| A C \\

\mathrm{PR}=\frac{1}{2} A C=\frac{21}{2} \mathrm{~cm}\\

Perimeter of quadrilateral A R P Q=A R+P R+Q P+A Q \\

=(\frac{21}{2}+\frac{21}{2}+15+15) \mathrm{cm}=51 \mathrm{~cm}(\text { Ans })\\

(iii) In the triangle ABC, D is any point on the side A C. The mid points of AB, BC, AD and D C are P, Q, X, Y respectively. If P X = 5 cm. then let us write the length of the side QY.

Solution:

In \triangle A B D P is mid point of AB and X is midpoint of AD

\therefore P X \| B D \\

\therefore P X=\frac{1}{2} B D---(i)\\

Again, In \triangle B C D \\

Y is mid point of CD and Q is midpoint BC

\therefore Q Y \| B D \\

\therefore Q Y=\frac{1}{2} B D------(ii)\\

From (i) \& (ii) P X=Q Y \\

\therefore Q Y=5 \mathrm{~cm}(\text { Ans })\\

(iv) In the triangle ABC, the medians BE and CF intersects at the point G. The midpoints of BG and CG are P and Q respectively. If P Q = 3 cm then let us write the length of BC.

Solution:

In \triangle B C G P is mid point of BG and Q is mid point of GC

\therefore P Q \| B C \\

\therefore P Q=\frac{1}{2} B C \\

= 2 \times 3 \mathrm{~cm} =6 \mathrm{~cm} (Ans)\\

(v) In the triangle A B C, the midpoints of BC, CA and AB are D, E and F respectively; FE intersects AD at the point O. If AD = 6 cm .let us write the length of AO.

Solution:

In \triangle A B C \\

F is mid point of AB & D is mid point of BC

\therefore F D \| A C \text { i.e. } F D \| A E\\

Again, E is midpoint of A C & D is mid point of BC

\therefore D E \| A B \text { ie } D E \| A F\\

Hence F D\|A E \ L E\| A F \\

\therefore AEDF is a parallelogram.

Since diagonal of a parallelogram bisect each other

\therefore A O=O D \\

\therefore A O+O D=A D \\

\text { or, } A O+A O=6 \\

\text { or, } 2 A O=6 \\

\therefore A O=3 \mathrm{~cm} \text { (Ans) }\\

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