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)\\

Adding (i) and (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.

**16. Short-answer type question:**

**(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) }\\