The availability of step-by-step ICSE Class 9 Maths Solutions S Chand Chapter 9 Mid-Point and Intercept Theorems Ex 9(A) can make challenging problems more manageable.

## S Chand Class 9 ICSE Maths Solutions Chapter 9 Mid-Point and Intercept Theorems Ex 9(A)

Question 1.

Fill in the blanks :

(i) The line joining the mid-points of two sides of a triangle is …………. to the third side.

(ii) The line drawn through the mid-point of one side of a triangle parallel to another side bisects the …………. side.

(iii) In figure, S and T are the mid-points of PQ and PR respectively. If ST = 3 cm, then QR = ………….

(iv) In figure, D and E are the mid-points of AB and AC. If DE = 7.5 cm, then BC = ………….

(v) Find the value of x in each triangle.

Solution:

(i) The line joining the mid-points of two sides of a triangle is parallel to the third side.

(ii) The line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.

(iii) In figure, S and T are the mid-points of PQ and PR respectively. If ST = 3 cm, then QR = 6 cm.

(iv) In figure, D and E are the mid-points of AB and AC. If DE = 7.5 cm, then BC = 15 cm.

(v) (a) In the given figure,

In ∆ABC

AD = DB and CE = EB

D and E are the mid points of AB and BC respectively

∴ DE || AC

∴ DE = \(\frac { 1 }{ 2 }\) AC and DE || AC

64 = \(\frac { 1 }{ 2 }\) x 4x

⇒ 2x = 64 ⇒ x = \(\frac { 64 }{ 2 }\) = 32

(b) In ∆ABC,

AD = DB and AE = EC

∴ DE || BC and

DE = \(\frac { 1 }{ 2 }\)BC

⇒ x – 8 = \(\frac { 1 }{ 2 }\) x 40 ⇒ x – 8 = 20

⇒ x = 20 + 8 = 28

∴ x = 28

(c) In ∆ABC,

AD = DC and BE = EC

∴ DE || AB and DE = \(\frac { 1 }{ 2 }\) AB

x + 8 = \(\frac { 1 }{ 2 }\) x 6x ⇒ x + 8 = 3x ⇒ 3x – x = 8

⇒ 2x = 8 ⇒ x = \(\frac { 8 }{ 2 }\) = 4

∴ x = 4

Question 2.

In ∆ABC, AB = 6 cm, and AC = 3 cm. If M is the mid-point of AB, and a straight line through M parallel to AC cuts BC in N, what is the length of MN?

Solution:

In ∆ABC, AB = 6 cm, AC = 3 cm

M is mid point of AB

A line through M is drawn parallel to AC which cuts BC in N

∵ M is mid point of AB and MN || AC

∴ N is mid point of BC

∴ MN = \(\frac { 1 }{ 2 }\) AC

= \(\frac { 1 }{ 2 }\) x 3 cm = 1.5 cm

Question 3.

In the figure, in the ∆ABC, D is the mid-point of AB, E is the mid-point of AC. Calculate:

(i) DE, if BC = 6 cm

(ii) ∠ADE, if ∠DBC = 140°

Solution:

In the figure,

∆ABC, D is mid-point of AB and E is mid-point of AC

∴ DE || BC and DE = \(\frac { 1 }{ 2 }\) BC

(i) ∵ BC = 6 cm

∴ DE = \(\frac { 1 }{ 2 }\) x 6 = 3 cm

(ii) ∵ ∠DBC = 140°

∵ DE || BC

∴ ∠ADE = ∠DBC or ∠ABC

(corresponding angles)

∴ ADE = 140°

Question 4.

In figure, D, E, F are the mid-points of BC, CA and AB respectively. Prove that AD bisects EF.

Solution:

Given : D, E and F are the mid-points of the sides BC, CA and AB respectively AD and EF are joined

To prove : AD bisects EF

Construction : Join ED and FD

Proof: In ∆ABC

E and F are the mid-points of CA and AB 1

∴ EF || BC and EF = \(\frac { 1 }{ 2 }\) BC

Similarly we can prove that

ED || AB and equal to \(\frac { 1 }{ 2 }\) AB

and FD || AC and equal to \(\frac { 1 }{ 2 }\) AC

∴ ∆EDF is a parallelogram

∵ The diagonals bisect each other

∴ AD bisects EF

Question 5.

In figure, ABC is an isosceles triangle in which AB = AC. D, E and F are respectively the mid-points of sides BC, AB and CA. Show that the line segments AD, EF bisect each other at right angles.

Solution:

Given : ∆BC is an isosceles triangle in which AB = AC

D, E and F are the mid-points of BC, AB and AC respectively

AD and EF are joined

To prove : AD and EF are perpendicular to each other

Proof: ∵ E and F are mid-points of AB and AC

∴ AE = EB and AF = FC

But AB = AC (given)

∴ AE = AF

and EF || BC and EF = \(\frac { 1 }{ 2 }\) BC

Now in ∆ABD and ∆ACD

AB = AC (given)

BD = DC (D is mid-point)

AD = AD (common)

∴ ∆ABD = ∆ACD (SSS axiom)

∴ ∠BAD = ∠CAD (c.p.c.t)

Now in ∆AEG and ∆AFG

AG = AG (common)

AE = AF (\(\frac { 1 }{ 2 }\) of equal sides)

∠BAD or ∠EAG = ∠CAD or ∠AFG (proved)

∴ ∆AEG ≅ ∆AFG (SAS axiom)

∴ ∠AGE = ∠AGF (c.p.c.t.)

But ∠AGE + ∠AFG = 180° (Linear pair)

∴ ∠AGE = ∠AGF = 90°

∴ AD and EF are perpendicular to each other

Hence proved.

Question 6.

Prove that the four triangles formed by joining the pairs of mid-points of the three sides of a triangle are congurent to each other.

Solution:

Given : In ∆ABC, D, E and F are the mid-points of the sides BC, CA and AB respectively

DE, EF and FD are joined

To prove : ∆AEF ≅ ∆BDF

∆DEF ≅ ∆CDE

Proof: ∵ D and E are the mid-points of side BC and CA respectively

∴ DE || AB and DE = \(\frac { 1 }{ 2 }\) AB

∴ DE = BF = FA … (i)

Similarly, D and F are the mid-points of BC andAB respectively

∴ DF || AC and DF = –\(\frac { 1 }{ 2 }\) AC … (ii)

DF = AE = EC

Now in ∆AEF and ∆DEF

AF = DE (proved)

AE = DF (proved)

EF = EF (common)

∴ ∆AEF ≅ ∆DEF … (i) (SSS criteria)

Similarly we can prove that

∆BDF ≅ ∆DEF … (i)

and ∆CDE ≅ ∆DEF … (ii)

From (i), (ii) and (iii)

∆AEF ≅ ∆BDF ≅ ∆CDE ≅ ∆DEF

Hence proved.

Question 7.

D, E and F are respectively the mid-points of the sides BC, CA and AB of an equilateral triangle ABC. Prove that ∆DEF is also an equilateral triangle.

Solution:

Given : ∆ABC is an equilateral triangle. D, E and F are the mid-points of sides BC, CA and AB respectively

DE, EF and FD are joined

To prove : ∆DEF is an equilateral triangle

Proof:

∵ E and F are the mid-points of AC and AB respectively

∴ EF || BC and EF = \(\frac { 1 }{ 2 }\) BC … (i)

Similarly, D and E are the mid-points of BC and AC respectively

∴ DE || AB and DE = \(\frac { 1 }{ 2 }\) AB … (ii)

and D and F are the mid-points of BC and AB respectively

∴ DF || AC and DF = \(\frac { 1 }{ 2 }\) AC … (iii)

From (i), (ii) and (iii)

∵ AB = BC = CA (sides of equilateral ∆ABC)

∴\(\frac { 1 }{ 2 }\) EF = \(\frac { 1 }{ 2 }\) DE = \(\frac { 1 }{ 2 }\) DF

⇒ DE = EF = FD

∴ ∆DEF is an equilateral triangle

Hence proved.

Question 8.

In figure, AD and BE are the medians of ∆ABC and BE || DF. Prove that CF = \(\frac { 1 }{ 4 }\) AC.

Solution:

Given: In AABC, AD and BE are its median BE || DF

To prove : CF = \(\frac { 1 }{ 4 }\) AC

Proof:

∵ BE is the median

∴ E is mid-point of AC

∴ AE = EC or EC = \(\frac { 1 }{ 2 }\) AC … (i)

In ∆BCE,

D is mid-point of BC and DF || BE

∴ F is mid-point of EC

∴ FC = \(\frac { 1 }{ 2 }\) EC … (iv)

From (i) and (ii),

CF = \(\frac { 1 }{ 2 }\) EC = \(\frac { 1 }{ 2 }\) (\(\frac { 1 }{ 2 }\) AC) = \(\frac { 1 }{ 4 }\)AC

Hence CF = \(\frac { 1 }{ 4 }\) AC

Hence proved.

Question 9.

In the given figure, ABC is an isosceles A with AB = AC and CP || BA and AP is the bisector of ext. ∠CAD of ∆ABC. Prove that

(i) ∠PAC = ∠BCA

(ii) ∆BCP is a parallelogram.

Solution:

Given : ∆ABC is an isosceles triangle in which AB = AC

CP || BA is drawn and AP is the bisector of ∠CAD

To prove :

(i) ∠PAC = ∠BCA

(ii) ABCP is a parallelogram Proof: In ∆ABC,

(i) ∵ AB=AC (given)

∴ ∠C = ∠B (Angles opposite to equal sides)

But Ext. ∠CAD = ∠B + ∠C= 2∠C … (i)

(Sum of its interior opposite angles)

and ∠PAC = ∠PAD = \(\frac { 1 }{ 2 }\) ∠CAP … (ii)

From (i) and (ii),

∠C = ∠CAP or ∠PAC = ∠BCA

(ii) ∵∠PAC = ∠BCA (proved)

But these are alternate angles

∴ AP || BC

But CP || BA

∴ ABCP is a parallelogram

Hence proved.

Question 10.

ABCD is a kite having AB = AD and BC = CD. Prove that the figure formed by joining the midpoints of the sides, in order, is a rectangle.

Solution:

Given: In the figure,ABCD is a kite in which AB = AD and BC = DC

P, Q, R and S are the mid-points of the sides AB, AD, DC and BC respectively

PQ, QR, RS and PS are joined

To prove : PQRS is a rectangle

Construction : Join BD and AC

Proof: In ∆ABD,

∵ P and Q are mid-points of AB and AD respectively

∴ PQ || BD and PQ = \(\frac { 1 }{ 2 }\) BD … (i)

Similarly in ABCD,

S and R are the mid-points of BC and DC

∴ SR || BD and SR = \(\frac { 1 }{ 2 }\) BD … (ii)

From (i) and (ii)

∴ PQ = SR and PQ || SR

Similarly PS = QR and PS || QR

∴ PQRS is a parallelogram

∵ Diagonals AC and BD intersect

Each other at right angles

∴ PQRS is a rectangle

Hence proved.

Question 11.

Show that the quadrilateral, fonned by joining the mid-points of the consecutive sides of a rectangle, is a rhombus.

Solution:

Given : ∆BCD is a rectangle. P, Q, R, and S are the mid-points of the sides AB, BC, CD and DA respectively PQ, QR, RS and SP are joined

To prove : PQRS is a rhombus

Construction : Join PR and SQ, AC and BD are also joined

Proof: In ∆ABC,

∵ P and Q are mid-points of AB and BC respectively

∴ PQ || AC and PQ = \(\frac { 1 }{ 2 }\) AC … (i)

Similarly in ∆ADC,

∴ S and R are the mid-points of AD and CD respectively

∴ SR || AC and SR = \(\frac { 1 }{ 2 }\)AC … (ii)

From (i) and (ii)

PQRS is a parallelogram

Similarly PS || BD and SP = \(\frac { 1 }{ 2 }\) BD 1

and QR || BD and QR = \(\frac { 1 }{ 2 }\) BD

But in rectangle ABCD diagonals AC and BD are equal

∴ PQRS is a rhombus

Hence proved.

Question 12.

Quad. ABCD is a rhombus and P, Q, R, S are the mid-points of AB, BC, CD, DA respectively. Prove that quad. PQRS is a rectangle.

Solution:

Given : ABCD is a quad, which is a rhombus. P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively PQ, QR, RS and SP are joined

To prove : PQRS is a rectangle

Construction : Join AC and BD

Proof: In ∆ABC

∵ P and Q are the mid-points of AB and CD respectively

∴ PQ || AC and PQ = \(\frac { 1 }{ 2 }\) AC … (i)

Similarly in ∆ADC,

R and S are the mid-points of CD and DA respectively

∴ RS || AC and RS = \(\frac { 1 }{ 2 }\) AC … (ii)

∴ PQRS is a parallelogram

∵ Diagonals of rhombus bisect each other at right angles PQRS is a rectangle

Hence proved.

Question 13.

In the quadrilateral ABCD the mid-points of AB, BC, CD, DA are L, M, P, Q.

(i) Using the mid-point theorem make a statement concerning the lengths and directions of LM and AC.

(ii) Prove that LMPQ is a parallelogram.

(iii) If it is also given that the diagonals AC and BD are equal. What further statement can be made about the parallelogram LMPQ ?

Solution:

Given : In quadrilateral ABCD, L, M, Pand Q are the mid-points of the sides AB, BC, CD and DA respectively. LM and AC are joined.

To prove :

(i) Using the mid point theorem, make a statement concerning the lengths and directions of LM and AC

(ii) Prove that LMPQ is a parallelogram.

(iii) If it is also given that the diagonals AC and BD are equal what further statement can be made about the parallelogram LMPQ ?

Construction : Join BD, MP, PQ and QL

Proof:

(i) ∵ L and M are the mid-points of AB and BC respectively

∴ LM || AC and LM = \(\frac { 1 }{ 2 }\) AC … (i)

(ii) Similarly P and Q are the mid-points of CD and DA respectively

∴ PQ || AC and PQ = \(\frac { 1 }{ 2 }\) AC … (ii)

From (i) and (ii),

LMPQ is a parallelogram

(iii) If diagonals AC and BD are equal

Then ABCD can be a parallelogram or a rectangle

Then LM = MP = PQ = QL Then LMPQ will be a rhombus

(∵ If the sides of a parallelogram are equal then it is a rhombus)

Question 14.

ABCD is a parallelogram E is the mid-point of AB and F is the mid-point of CD. GH is any line that intersects AD, EF and BC in G, P and H respectively. Prove that GP = PH.

Solution:

Given : In the figure ABCD is a parallelogram. E and F are the mid-points of AB and CD respectively GH is any line which intersects AD in G, EF in P and BC in H

To prove : GP = PH

Proof:

∵ E and F are the mid-points of AB and DC respectively

∴ EF || AD || BC (∵ ABCD is a parallelogram)

∴ AE = EB ⇒ \(\frac { AE }{ EB }\) = 1

∴ \(\frac { GP }{ PH }\) = 1 ⇒ GP = PH (intercept theorem)

Hence proved.

Question 15.

Prove that in a parallelogram, the lines joining a pair of opposite vertices to the mid-points of a pair of opposite sides trisect a diagonal.

Solution:

Given : In ||gm ABCD, E and F are the mid-points of side AD and BC respectively

EB and DF are joined which intersect diagonal AC at M and N

To prove : M and N trisect AC

i.e. AM = MN = NC

Proof: ∵ E and F are the mid-points of AD and BC respectively

∴ BE || DF

Now in ∆ADN,

∵ EM || DN and AE = ED

AM = MN … (i)

Similarly in ∆BCM,

∵ CF = FB

∴ CN = NM … (ii)

From (i) and (ii),

AM = MN = NC

Hence M and N trisect AC

Hence proved.

Question 16.

In figure, E is the mid-point of side AD of a trapezium ABCD, with AB || DC. A line through E parallel to AB, meets BC in F. Show that F is the mid-point of BC.

Solution:

Given : In trapezium ABCD, AB || DC

E is mid-point of AD

A line through E is drawn parallel to AB which meets BC in F

To prove : F is the mid-point of BC

Proof: In ∆ADB,

∵ E is mid-point of AD and EO orEF is parallel toAB

∴ O is mid-point of BD

Similarly in ∆BCD,

O is mid-point of BD and EOF || CD (||AB)

∴ F is mid-point BC

Hence proved.