# ML Aggarwal Solutions for Chapter 11 Mid Point Theorem Class 9 Maths ICSE

**Exercise 11**

**1. (a) In the figure (1) given below, D, E and F are mid-points of the sides BC, CA and AB respectively of Î” ABC. If AB = 6 cm, BC = 4.8 cm and CA = 5.6 cm, find the perimeter of (i) the trapezium FBCE (ii) the triangle DEF.**

**(b) In the figure (2) given below, D and E are mid-points of the sides AB and AC respectively. If BC = 5.6 cm and ∠B = 72°, compute (i) DE (ii) ∠ADE.**

**(c) In the figure (3) given below, D and E are mid-points of AB, BC respectively and DF || BC. Prove that DBEF is a parallelogram. Calculate AC if AF = 2.6 cm.**

**Solution**

**(a) (i)** It is given that

AB = 6 cm, BC = 4.8 cm and CA = 5.6 cm

To find: The perimeter of trapezium FBCA

It is given that

F is the mid-point of AB

We know that

BF = ½ AB = ½ × 6 = 3 cm **…(1)**

It is given that

E is the mid-point of AC

We know that

CE = ½ AC = ½ × 5.6 = 2.8 cm **…(2)**

Here, F and E are the mid-point of AB and CA

FE || BC

We know that

FE = ½ BC = ½ × 4.8 = 2.4 cm **…(3)**

Here

Perimeter of trapezium FBCE = BF + BC + CE + EF

Now substituting the value from all the equations

= 3 + 4.8 + 2.8 + 2.4

= 13 cm

Therefore, the perimeter of trapezium FBCE is 13 cm.

**(ii) **D, E and F are the midpoints of sides BC, CA and AB of Î” ABC

Here EF || BC

EF = ½ BC = ½ × 4.8 = 2.4 cm

DE = ½ AB = ½ × 6 = 3 cm

FD = ½ AC = ½ × 5.6 = 2.8 cm

We know that

Perimeter of Î” DEF = DE + EF + FD

Substituting the values

= 3 + 2.4 + 2.8

= 8.2 cm

**(b)** It is given that

D and E are the mid-point of sides AB and AC

BC = 5.6 cm and ∠B = 72°

To find: (i) DE (ii) ∠ADE

We know that

In Î” ABC

D and E is the mid-point of the sides AB and AC

Using mid-point theorem

DE || BC

**(i)** DE = ½ BC = ½ × 5.6 = 2.8 cm

**(ii)** ∠ADE = ∠B are corresponding angles

It is given that

∠B = 72° and BC || DE

∠ADE = 72°

**(c)** It is given that

D and E are the midpoints of AB and BC respectively

DF || BC and AF = 2.6 cm

To find: (i) BEF is a parallelogram

**(ii)** Calculate the value of AC

Proof:

**(i)** In Î” ABC

D is the midpoint of AB and DF || BC

F is the midpoint of AC **…(1)**

F and E are the midpoints of AC and BC

EF || AB **…(2)**

Here DF || BC

DF || BE** …(3)**

Using equation (2)

EF || AB

EF || DB **…(4)**

Using equation (3) and (4)

DBEF is a parallelogram

(ii) F is the midpoint of AC

So we get

AC = 2×AF = 2× 2.6 = 5.2 cm

**2. Prove that the four triangles formed by joining in pairs the mid-points of the sides C of a triangle are congruent to each other.**

**Solution**

It is given that

In Î” ABC

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

Now join DE, EF and FD

To find:

Î” ADF ≅ Î” DBE ≅ Î” ECF ≅ Î” DEF

To prove:

In Î” ABC

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

DE || AC or FC

Similarly DF || EC

DECF is a parallelogram

We know that

Diagonal FE divides the parallelogram DECF in two congruent triangles DEF and CEF

Î” DEF ≅ Î” ECF **...(1)**

Here we can prove that

Î” DBE ≅ Î” DEF **…(2)**

Î” DEF ≅ Î” ADF **...(3)**

Using equation (1), (2) and (3)

Î” ADF ≅ Î” DBE ≅ Î” ECF ≅ Î” DEF

**3. If D, E and F are mid-points of sides AB, BC and CA respectively of an isosceles triangle ABC, prove that Î” DEF is also isosceles.**

**Solution**

It is given that

ABC is an isosceles triangle in which AB = AC

D, E and F are the midpoints of the sides BC, CA and AB

Now D, E and F are joined

To find:

Î” DEF is an isosceles triangle

Proof:

D and E are the midpoints of BC and AC

Here DE || AB and DE = ½ AB **…(1)**

D and F are the midpoints of BC and AB

Here DF || AC and DF = ½ AC **…(2)**

It is given that

AB = BC and DE = DF

Hence, Î” DEF is an isosceles triangle.

**4. The diagonals AC and BD of a parallelogram ABCD intersect at O. If P is the midpoint of AD, prove that**

**(i) PQ || AB**

**(ii) PO = ½ CD.**

**Solution**

It is given that

ABCD is a parallelogram in which diagonals AC and BD intersect each other

At the point O, P is the midpoint of AD

Join OP

To find: (i) PQ || AB (ii) PQ = ½ CD

**Proof:**

**(i) **In parallelogram diagonals bisect each other

BO = OD

Here O is the mid-point of BD

In Î”ABD,

P and O is the midpoint of AD and BD

PO || AB and PO = ½ AB **…(1)**

Hence, it is proved that PO || AB.

**(ii)** ABCD is a parallelogram

AB = CD **…(2)**

Using both (1) and (2)

PO = ½ CD

**5. In the adjoining figure, ABCD is a quadrilateral in which P, Q, R and S are mid-points of AB, BC, CD and DA respectively. AC is its diagonal. Show that**

**(i) SR || AC and SR = ½ AC**

**(ii) PQ = SR**

**(iii) PQRS is a parallelogram.**

**Solution**

It is given that

In quadrilateral ABCD

P, Q, R and S are the mid-points of sides AB, BC, CD and DA

AC is the diagonal

To find:

(i) SR || AC and SR = ½ AC

(ii) PQ = SR

(iii) PQRS is a parallelogram

**Proof:**

**(i)** In Î” ADC

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

SR || AC and SR = ½ AC** …(1) [Using the mid-point theorem]**

**(ii)** In Î” ABC

P and Q are the midpoints of AB and BC

PQ || AC and PQ = ½ AC **…(2)**

Using equation (1) and (2)

PQ = SR and PQ || SR

**(iii)** PQ = SR and PQ || SR

Hence, PQRS is a parallelogram.

**6. Show that the quadrilateral formed by joining the mid-points of the adjacent sides of a square, is also a square.**

**Solution**

It is given that

A square ABCD in which E, F, G and H are mid-points of AB, BC, CD and DA

Join EF, FG, GH and HE.

To find:

EFGH is a square

Construct AC and BD

**Proof:**

In Î” ACD

G and H are the mid-points of CD and AC

GH || AC and GH = ½ AC **…(1)**

In Î” ABC, E and F are the mid-points of AB and BC

EF || AC and EF = ½ AC **…(2)**

Using both the equations

EF || GH and EF = GH = ½ AC **…(3)**

In the same way we can prove that

EF || GH and EH = GF = ½ BD

We know that the diagonals of square are equal

AC = BD

By dividing both sides by 2

½ AC = ½ BD **…(4)**

Using equation (3) and (4)

EF = GH = EH = GF **…(5)**

Therefore, EFGH is a parallelogram

In Î” GOH and Î” GOF

OH = OF as the diagonals of parallelogram bisect each other

OG = OG is common

**Using equation (5)**

GH = GF

Î” GOH ≅ Î” GOF **(SSS axiom of congruency)**

∠GOH = ∠GOF **(c.p.c.t)**

We know that

∠GOH + ∠GOF = 180° as it is a linear pair

∠GOH + ∠GOH = 180°

So we get

2 ∠GOH = 180^{0}

∠GOH = 180°/2 = 90°

So the diagonals of a parallelogram ABCD bisect and perpendicular to each other

Hence, it is proved that EFGH is a square.

**7. In the adjoining figure, AD and BE are medians of Î” ABC. If DF || BE, prove that CF = ¼ AC.**

**Solution**

It is given that

AD and BE are the medians of Î” ABC

Construct DF || BE

To find:

CF = ¼ AC

**Proof:**

In Î” BCE

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

F is the mid-point of EC

CF = ½ EC **…(1)**

E is the mid-point of AC

EC = ½ AC **…(2)**

Using both the equations

CF = ½ EC = ½ (½ AC)

So we get,

CF = ¼ AC

Hence, it is proved.

**8. In the figure (1) given below, ABCD is a parallelogram. E and F are mid-points of the sides AB and CO respectively. The straight lines AF and BF meet the straight lines ED and EC in points G and H respectively. Prove that**

**(i) Î” HEB = Î” HCF**

**(ii) GEHF is a parallelogram.**

**Solution**

It is given that

ABCD is a parallelogram

E and F are the mid-points of sides AB and CD

To prove:

(i) Î” HEB = Î” HCF

(ii) GEHF is a parallelogram

**Proof:**

**(i)** We know that

ABCD is a parallelogram

FC || BE

∠CEB = ∠FCE are alternate angles

∠HEB = ∠FCH **…(1)**

∠EBF = ∠CFB are alternate angles

∠EBH = ∠CFM **...(2)**

Here E and F are mid-points of AB and CD

BE = ½ AB **…(3)**

CF = ½ CD **...(4)**

We know that

ABCD is a parallelogram

AB = CD

Now dividing both sides by ½

½ AB = ½ CD

Using equation (3) and (4)

BE = CF **…(5)**

In Î” HEB and Î” HCF

∠HEB = ∠FCH **(using equation 1)**

∠EBH = ∠CFH **(using equation 2)**

BE = CF **(using equation 5)**

So we get,

Î” HEB ≅ Î” HCF **(ASA axiom of congruency)**

Hence, it is proved.

**(ii)** It is given that

E and F are the mid-points of AB and CD

AB = CD

So we get

AE = CF

Here AE || CF

AE = CF and AE || CF

So, AECF is a parallelogram.

G and H are the mid-points of AF and CE

GF || EH **…(6)**

In the same way we can prove that GFHE is a parallelogram

So G and H are the points on the line DE and BF

GE || HF** …(7)**

Using equation (6) and (7) GEHF is a parallelogram.

Hence, it is proved.

**9. ABC is an isosceles triangle with AB = AC. D, E and F are mid-points of the sides BC, AB and AC respectively. Prove that the line segment AD is perpendicular to EF and is bisected by it.**

**Solution**

It is given that

ABC is an isosceles triangle with AB = AC

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

To find:

AD is perpendicular to EF and is bisected by it.

**Proof:**

In Î” ABD and Î” ACD

ABC is an isosceles triangle

∠ABD = ∠ACD

Here D is the mid-point of BC

BD = BD

It is given that AB = AC

Î” ABD ≅ Î” ACD **(SAS axiom of congruency)**

∠ADB = ∠AOC **(c. p. c. t)**

We know that

∠ADB + ∠AOC = 180° is a linear pair

∠ADB + ∠ADB = 180°

By further calculation

2 ∠ADB = 180°

So, we get

∠ADB = 180°/2 = 90°

So AD is perpendicular to BC **…(1)**

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

DE || AF **…(2)**

D and F are the mid-points of BC and AC

EF || AD **…(3)**

Using equation (2) and (3)

AEDF is a parallelogram.

Here the diagonals of a parallelogram bisect each other

AD and EF bisect each other

Using equation (1) and (3)

EF || BC

So AD is perpendicular to EF

Hence, it is proved.

**10. (a) In the quadrilateral (1) given below, AB || DC, E and F are the mid-points of AD and BD respectively. Prove that:**

**(i) G is mid-point of BC**

**(ii) EG = ½ (AB + DC)**

**(b) In the quadrilateral (2) given below, AB || DC || EG. If E is mid-point of AD prove that:**

**(i) G is the mid-point of BC**

**(ii) 2EG = AB + CD**

**(c) In the quadrilateral (3) given below, AB || DC.**

**E and F are mid-point of non-parallel sides AD and BC respectively. Calculate:**

**(i) EF if AB = 6 cm and DC = 4 cm.**

**(ii) AB if DC = 8 cm and EF = 9 cm.**

**Solution**

**(a) **It is given that

AB || DC, E and F are mid-points of AD and BD

To prove:

(i) G is mid-point of BC

(ii) EG = ½ (AB + DC)

**Proof:**

**(i)** In Î” ABD

F is the mid-point of BD

DF = BF

E is the mid-point of AD

EF || AB and EF = ½ AB **…(1)**

It is given that AB || CD

EG || CD

F is the mid-point of BD

FG || DC

G is the mid-point of BC

**(ii)** FG = ½ DC** …(2)**

By adding both the equations

EF + FG = ½ AB + ½ DC

Taking ½ as common

EG = ½ (AB + DC)

Therefore, it is proved.

**(b)** It is given that

Quadrilateral ABCD in which AB || DC || EG

E is the mid-point of AD

To prove:

(i) G is the mid-point of BC

(ii) 2EG = AB + CD

**Proof:**

**(i)** AB || DC

EG || AB

So we get

EG || DC

In Î” DAB,

E is the mid-point of BD and EF = ½ AB **…(1)**

In Î” BCD,

F is the mid-point of BD and FG || DC

FG = ½ CD **…(2)**

By adding both the equations

EF + FG = ½ AB + ½ CD

Taking out the common terms

EG = ½ (AB + CD)

Hence, it is proved.

**(c)** It is given that

A quadrilateral in which AB || DC

E and F are the mid-points of non-parallel sides AD and BC

To prove:

(i) EF if AB = 6 cm and DC = 4 cm.

(ii) AB if DC = 8 cm and EF = 9 cm.

**Proof:**

We know that

The length of line segment joining the mid-points of two non-parallel sides is half the sum of the lengths of the parallel sides

E and F are the mid-points of AD and BC

EF = ½ (AB + CD) **...(1)**

**(i) **AB = 6 cm and DC = 4 cm

Substituting in equation (1)

EF = ½ (6 + 4)

By further calculation

EF = ½ × 10 = 5 cm

**(ii)** DC = 8 cm and EF = 9 cm

Substituting in equation (1)

EF = ½ (AB + DC)

By further calculation

9 = ½ (AB + 8)

⇒ 18 = AB + 8

So we get

18 – 8 = AB

⇒ AB = 10 cm

**11. (a) In the quadrilateral (1) given below, AD = BC, P, Q, R and S are mid-points of AB, BD, CD and AC respectively. Prove that PQRS is a rhombus.**

**(b) In the figure (2) given below, ABCD is a kite in which BC = CD, AB = AD, E, F, G are mid-points of CD, BC and AB respectively. Prove that:**

**(i) ∠EFG = 90°**

**(ii) The line drawn through G and parallel to FE bisects DA.**

**Solution**

**(a)** It is given that

A quadrilateral ABCD in which AD = C

P, Q, R and S are mid-points of AB, BD, CD and AC

To prove:

PQRS is a rhombus

**Proof:**

**(i)** In Î” ABD

P and Q are mid points of AB and BD

PQ || AD and PQ = ½ AB **…(1)**

In Î” BCD,

R and Q are mid points of DC and BD

RQ || BC and RQ = ½ BC **...(2)**

P and S are mid-points of AB and AC

PS || BC and PS = ½ BC **…(3)**

AD = BC

Using all the equations

PS || RQ and PQ = PS = RQ

Here, PS || RQ and PS = RQ

PQRS is a parallelogram

PQ = RS = PS = RQ

PQRS is a rhombus

Therefore, it is proved.

**(ii)** It is given that

ABCD is a kite in which BC = CD, AB = AD, E, F, G are mid-points of CD, BC and AB

To prove:

(i) ∠EFG = 90°

(ii) The line drawn through G and parallel to FE bisects DA

Construction:

Join AC and BD

Construct GH through G parallel to FE

**Proof:**

**(i)** We know that

Diagonals of a kite interest at right angles

∠MON = 90° **…(1)**

In Î” BCD

E and F are mid-points of CD and BC

EF || DB and EF = ½ DB **...(2)**

EF || DB

MF || ON

Here

∠MON + ∠MFN = 180°

⇒ 90° + ∠MFN = 180°

By further calculation

∠MFN = 180° – 90° = 90°

So, ∠EFG = 90°

Hence, it is proved.

**(ii)** In Î” ABD

G is the mid-point of AB and HG || DB

Using equation (2)

EF || DB and EF || HG

HG || DB

Here, H is the mid-point of DA

Therefore, the line drawn through G and parallel to FE bisects DA.

**12. In the adjoining figure, the lines l, m and n are parallel to each other, and G is mid-point of CD. Calculate:**

**(i) BG if AD = 6 cm**

**(ii) CF if GE = 2.3 cm**

**(iii) AB if BC = 2.4 cm**

**(iv) ED if FD = 4.4 cm**

**Solution**

It is given that

The straight line l, m and n are parallel to each other

G is the mid-point of CD

To find:

(i) BG if AD = 6 cm

(ii) CF if GE = 2.3 cm

(iii) AB if BC = 2.4 cm

(iv) ED if FD = 4.4 cm

**Proof:**

**(i)** In Î” ACD,

G is the mid-point of CD

BG || AD as m || n

Here B is the mid-point of AC and BG = ½ AD

So we get

BG = ½ × 6 = 3 cm

**(ii)** In Î” CDF

G is the mid-point of CD

GE || CF as m || l

Here E is the mid-point of DF and GE = ½ CF

So we get

CF = 2GE

CF = 2 ×2.3 = 4.6 cm

**(iii)** From (i)

B is the mid-point of AC

AB = BC

We know that

BC = 2.4 cm

So AB = 2.4 cm

**(iv)** From (ii)

E is the mid-point of FD

ED = ½ FD

We know that

FD = 4.4 cm

⇒ ED = ½ × 4.4 = 2.2 cm

**Chapter Test**

**1. ABCD is a rhombus with P, Q and R as midpoints of AB, BC and CD respectively. Prove that PQ ⊥QR.**

**Solution**

It is given that

ABCD is a rhombus with P, Q and R as mid-points of AB, BC and CD

To prove:

PQ ⊥QR

Construction: Join AC and BD

**Proof:**

Diagonals of rhombus intersect at right angle

∠MON = 90° **…(1)**

In Î” BCD

Q and R are mid-points of BC and CD.

RQ || DB and RQ = ½ DB **…(2)**

Here

RQ || DB

MQ || ON

We know that

∠MQN + ∠MON = 180°

Substituting the values

∠MQN + 90° = 180°

⇒ ∠MQN = 180° – 90° = 90°

So, NQ ⊥MQ or PQ ⊥QR

Hence, it is proved.

**2. The diagonals of a quadrilateral ABCD are perpendicular. Show that the quadrilateral formed by joining the mid-points of its adjacent sides is a rectangle.**

**Solution**

It is given that

ABCD is a quadrilateral in which diagonals AC and BD are perpendicular to each other

P, Q, R and S are mid-points of AB, BC, CD and DA

To prove:

PQRS is a rectangle

**Proof:**

We know that

P and Q are the mid-points of AB and BC

PQ || AC and PQ = ½ AC **…(1)**

S and R are mid-points of AD and DC

SR || AC and SR = ½ AC **…(2)**

Using both the equations

PQ || SR and PQ = SR

So PQRS is a parallelogram

AC and BD intersect at right angles

SP || BD and BD ⊥AC

So SP ⊥ AC i.e. SP ⊥SR

∠RSP = 90°

⇒ ∠RSP = ∠SRQ = ∠RQS = ∠SPQ = 90°

Hence, PQRS is a rectangle.

**3. If D, E, F are mid-points of the sides BC, CA and AB respectively of a Î” ABC, prove that AD and FE bisect each other.**

**Solution**

It is given that

D, E, F are mid-points of sides BC, CA and AB of a Î” ABC

To prove:

AD and FE bisect each other

Construction:

Join ED and FD

**Proof:**

We know that

D and E are the midpoints of BC and AB

DE || AC and DE || AF **...(1)**

D and F are the midpoints of BC and AC

DF || AB and DF || AE **…(2)**

Using both the equations

ADEF is a parallelogram

Here the diagonals of a parallelogram bisect each other

AD and EF bisect each other.

Therefore, it is proved.

**4. In Î” ABC, D and E are mid-points of the sides AB and AC respectively. Through E, a straight line is drawn parallel to AB to meet BC at F. Prove that BDEF is a parallelogram. If AB = 8 cm and BC = 9 cm, find the perimeter of the parallelogram BDEF.**

**Solution**

It is given that

In Î” ABC

D and E are the mid points of sides AB and AC

DE is joined from E

EF || AB is drawn AB = 8 cm and BC = 9 cm

To prove:

(i) BDEI is a parallelogram

(ii) Find the perimeter of BDEF

**Proof:**

In Î” ABC

B and E are the mid-points of AB and AC

Here DE || BC and DE = ½ BC

So EF || AB

DEFB is a parallelogram

DE = BF

So we get

DE = ½ BC = ½ × 9 = 4.5 cm

EF = ½ AB = ½ × 8 = 4 cm

We know that

Perimeter of BDEF = 2 (DE + EF)

Substituting the values

= 2 (4.5 + 4)

= 2 ×8.5

= 17 cm

**5. In the given figure, ABCD is a parallelogram and E is mid-point of AD. DL || EB meets AB produced at F. Prove that B is mid-point of AF and EB = LF.**

**Solution**

It is given that

ABCD is a parallelogram

E is the mid-point of AD

DL || EB meets AB produced at F

To prove:

EB = LF

B is the mid-point of AF

**Proof:**

We know that

BC || AD and BE || LD

BEDL is a parallelogram

BE = LD and BL = AE

Here E is the mid-point of AD

L is the mid-point of BC

In Î” FAD

E is the mid-point of AD and BE || LD at FLD

So B is the mid-point of AF

Here, EB = ½ FD = LF

**6. In the given figure, ABCD is a parallelogram. If P and Q are mid-points of sides CD and BC respectively, show that CR = ½ AC.**

**Solution**

It is given that

ABCD is a parallelogram

P and Q are mid-points of CD and BC

To prove: CR = ¼ AC

Construction: Join AC and BD

Proof:

In parallelogram ABCD

Diagonals AC and BD bisect each other at O

AO = OC or OC = ½ AC **…(1)**

In Î” BCD

P and Q are mid points of CD and BC

PQ || BD

In Î” BCO,

Q is the mid-point of BC and PQ || OB

Here is the mid-point of CO

So, we get

CR = ½ OC = ½ (½ BC)

⇒ CR = ¼ BC

Hence, it is proved.