Frank Solutions for Class 9 Maths Chapter 15 Mid-point and Intercept Theorems

Frank Solutions for Class 9 Maths Chapter 15 Mid-point and Intercept Theorems act as an important study tool to prepare for Class 9 annual examinations. These solutions are prepared in accordance with the current ICSE Board syllabus, keeping in mind the understanding capacity of students. Practising these solutions on a daily basis will help them to achieve desired scores as well as enhance their problem-solving skills. Frank Solutions are provided here in PDF for students to practise offline without any time constraints and thus encourage concept-based learning.

Chapter 15 has problems related to the midpoints of a triangle. Mid-point theorem states that if any two sides of a triangle midpoints are connected, then the resulting line segment is parallel to the third side of the triangle. By referring to these solutions, students can cross-check their answers and clear doubts while solving textbook problems. Access Frank Solutions for Class 9 Maths Chapter 15 Mid-point and Intercept Theorems from the link below.

Frank Solutions for Class 9 Maths Chapter 15 Mid-point and Intercept Theorems Download PDF

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Access Frank Solutions for Class 9 Maths Chapter 15 Mid-point and Intercept Theorems

1. In △ABC, D is the mid-point of AB, and E is the mid-point of BC

FRANK Solutions Class 9 Maths Chapter 15 - 1

Calculate:

(i) DE, if AC = 8.6 cm

(ii) ∠DEB, if ∠ACB = 720

Solution:

In △ABC,

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

Hence, by mid-point theorem DE || AC and DE = (1/2) AC

(i) DE = (1/2) AC = (1/2) x 8.6

We get,

= 4.3 cm

(ii) ∠DEB = ∠C = 720 (Corresponding angles are equal, since DE || AC)

2. In △ABC, AB = 12 cm and AC = 9 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:

FRANK Solutions Class 9 Maths Chapter 15 - 2

MN || AC and M is the mid-point of AB

Hence, N is the mid-point of BC

Therefore, MN = (1/2) AC

= (9/2) cm

We get,

= 4.5 cm

3. (a) In △ABC, D, E, and F are the mid-points of BC, CA and AB, respectively. Find FE, if BC = 14 cm

(b) In △ABC, D, E, and F are the mid-points of BC, CA and AB, respectively. Find DE, if AB = 8 cm

(c) In △ABC, D, E, and F are the mid-points of BC, CA and AB, respectively. Fine ∠FDB if ∠ACB = 1150

Solution:

(a)

FRANK Solutions Class 9 Maths Chapter 15 - 3

F is the mid-point of AB, and E is the mid-point of AC

Hence,

FE = (1/2) BC ………… (Mid-point Theorem)

= (1/2) x 14

We get,

= 7 cm

(b)

FRANK Solutions Class 9 Maths Chapter 15 - 4

In △ABC,

D is the mid-point of BC, and E is the mid-point of AC

Hence,

DE = (1/2) AB …….. (Mid-point Theorem)

= (1/2) x 8

We get,

= 4 cm

(c)

FRANK Solutions Class 9 Maths Chapter 15 - 5

In △ABC,

FD || AC

Hence,

∠FDB = ∠ACB = 1150 …….(Corresponding angles are equal)

4. In parallelogram PQRS, L is the mid-point of side SR, and SN is drawn parallel to LQ, which meets RQ produced at N and cuts side PQ at M. Prove that M is the mid-point of PQ

FRANK Solutions Class 9 Maths Chapter 15 - 6

Solution:

In △NSR,

MQ = (1/2) SR

But L is the mid-point of SR and SR = PQ (Sides of parallelogram)

So, it can be written as,

MQ = (1/2) PQ

MQ = PM = LS= LR

Hence, M is the mid-point of PQ

5. In △ABC, BE and CF are medians. P is a point on BE produced such that BE = EP, and Q is a point on CF produced such that CF = FQ. Prove that:

(a) QAP is a straight line

(b) A is the mid-point of PQ

Solution:

FRANK Solutions Class 9 Maths Chapter 15 - 7

Since BE and CF are the medians,

F is the mid-point of AB, and E is the mid-point of AC

We know that the line joining the mid-points of any two sides is parallel and half of the third side

We have,

In △ACQ,

EF|| AQ and EF = (1/2) AQ ………(1)

In △ABP,

EF || AP and EF = (1/2) AP ……….(2)

(a) From (1) and (2)

We get,

AP || AQ (both are parallel to EF)

As AP and AQ are parallel and have a common point, A

This is possible only if QAP is a straight line

Thus, proved

(b) From (1) and (2),

EF = (1/2) AQ and EF = (1/2) AP

⇒ (1/2) AQ = (1/2) AP

 AQ = AP

Therefore, A is the mid-point of QP

6. Prove that the figure obtained by joining the mid-points of the adjacent sides of a rectangle is a rhombus

Solution:

FRANK Solutions Class 9 Maths Chapter 15 - 8

In the given rectangle ABCD,

Join AC and BD

In △ABC, P and Q are the mid-points of AB and BC, respectively

PQ = (1/2) AC …….. (1) and PQ ||AC

In △BDC, R and Q are the mid-points of CD and BC, respectively

QR = (1/2) BD ……..(2) and QR || BD

But AC and BD are diagonals of the rectangle

From equations (1) and (2)

PQ = QR

Similarly,

QR = RS and RS = SP

And,

RS || AC and SP || BD

Therefore, PQ = QR = RS = SP

Hence, PQRS is a rhombus

7. D, E and F are the mid-points of the sides AB, BC and CA of an isosceles △ABC in which AB = BC. Prove that △DEF is also isosceles.

Solution:

FRANK Solutions Class 9 Maths Chapter 15 - 9

E and F are mid-points of BC and AC

Hence, EF = (1/2) AB …….(1)

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

Hence, DF = (1/2) BC ……..(2)

But AB = BC

From (1) and (2)

EF = DF

Thus, △DEF is an isosceles triangle

8. The diagonals of a quadrilateral intersect each other at right angle. Prove that the figure obtained by joining the mid-points of the adjacent sides of the quadrilateral is a rectangle.

Solution:

FRANK Solutions Class 9 Maths Chapter 15 - 10

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

Hence,

PQ || AC and PQ = (1/2) AC ……. (i)

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

Hence,

SR || AC and SR = (1/2) AC …….(ii)

From (i) and (ii),

PQ || SR and PQ = SR

Therefore, PQRS is a parallelogram

Further, AC and BC intersect at right angles

∴ SP || BD and BD ⊥ AC

∴ SP ⊥ AC

⇒ SP ⊥ SR

⇒  ∠RSP = 900

∴ ∠RSP = ∠SRQ = ∠RQP = ∠SPQ = 900

Hence, PQRS is a rectangle

9. In a right-angled triangle ABC. ∠ABC = 900, and D is the mid-point of AC. Prove that BD = (1/2) AC

Solution:

FRANK Solutions Class 9 Maths Chapter 15 - 11

Draw line segment DE ||CB, which meets AB at point E

Now,

DE || CB and AB are the transversal,

∴∠AED = ∠ABC …. (corresponding angles)

∠ABC = 900 (given)

⇒ ∠AED = 900

Also,

Since D is the mid-point of AC and DE || CB,

DE bisects side AB,

Hence,

AE = BE …..(i)

In △AED and △BED,

∠AED = ∠BED …..(Each 900)

AE = BE ……[From (i)]

DE = DE ….(Common)

Therefore, △AED △BED ….(By SAS)

⇒ AD = BD …..(C.P.C.T.C)

⇒ BD = (1/2) AC

Hence, proved

10. In a parallelogram ABCD, E and F are the mid-points of the sides AB and CD, respectively. The line segments AF and BF meet the line segments DE and CE at points G and H, respectively. Prove that:

(a) △GEA ≅ △GFD

(b) △HEB ≅ △HFC

(c) EGFH is a parallelogram

Solution:

FRANK Solutions Class 9 Maths Chapter 15 - 12

Since ABCD is a parallelogram,

AB = CD and AD = BC

Now,

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

Hence,

AE = EB = DF = FC ……….(1)

(a) In △GEA and △GFD,

AE = DF …… [From (1)]

∠AGE = ∠DGF ……… (vertically opposite angles)

∠GAE = ∠GFD …….. (Alternate interior angles)

Therefore,

△GEA ≅ △GFD

(b) In △HEB and △HFC,

BE = FC …..[From (1)]

∠EHB = ∠FHC (vertically opposite angles)

∠HBE = ∠HFC (Alternate interior angles)

Therefore,

△HEB ≅ △HFC

(c) In quadrilateral AECF,

AE = CF …..[From (1)]

AE || CF ……(since AB || DC)

Hence,

AECF is a parallelogram

 EC || AF or EH || GF …… (i)

In quadrilateral BFDE,

BE = DF ….[From (1)]

BE || DF ….(since AB || DC)

⇒ BEDF is a parallelogram

 BF || ED or HF || EG ……(ii)

From equations (i) and (ii),

We get,

EGFH is a parallelogram

11. In △ABC, the medians BE and CD are produced to the points P and Q, respectively, such that BE = EP and CD = DQ. Prove that:

(a) Q, A and P are collinear

(b) A is the mid-point of PQ

Solution:

FRANK Solutions Class 9 Maths Chapter 15 - 13

In △BDC and △ADQ,

CD = DQ … (given)

∠BDC = ∠ADQ ……(vertically opposite angles)

BD = AD ……(D is the mid-point of AB)

Therefore,

△BDC ≅ △ADQ

 ∠DBC = ∠DAQ (cpct)…….(i)

And, BC = AQ (cpct) …….(ii)

Similarly,

We can prove

△CEB ≅ △ AEP

 ∠ECB = ∠EAP (cpct)……..(iii)

And, BC = AP (cpct)……..(iv)

(a) In △ABC,

∠ABC + ∠ACB + ∠BAC = 1800

 ∠DBC + ∠ECB + ∠BAC = 1800

 ∠DAQ + ∠EAP + ∠BAC = 1800 [From (i) and (iii)]

 Q, A, P are collinear

(b) From (ii) and (iv),

AQ = AP

Therefore, A is the mid-point of PQ

12. In △ABC, D and E are two points on the side AB such that AD = DE = EB. Through D and E, lines are drawn parallel to BC, which meet the side AC at points F and G, respectively. Through F and G, lines are drawn parallel to AB, which meets the side BC at points M and N, respectively. Prove that BM = MN = NC

FRANK Solutions Class 9 Maths Chapter 15 - 14

Solution:

FRANK Solutions Class 9 Maths Chapter 15 - 15

In △AEG,

D is the mid-point of AE and DF || EG

Hence,

F is the mid-point of AG

AF = FG ……….. (1)

In △ABC,

DF || EG|| BC

DE = BE

Hence,

GF = GC …………. (2)

From (1) and (2), we get,

AF = FG = GC

Similarly, since GN|| FM|| AB

Thus, BM = MN = NC (proved)

13. In the given figure, the lines l, m and n are parallel to each other. D is the mid-point of CE. Find

(a) BC

(b) EF

(c) CG

(d) BD

FRANK Solutions Class 9 Maths Chapter 15 - 16

Solution:

According to the equal intercept theorem,

Since CD = DE

AB = BC ………(i)

EF = GF ……….(ii)

(a) BC = AB = 6 cm ….. [From (i)]

(b) EG = EF + FG

EG = 2EF …… [From (ii)]

9 = 2EF

EF = (9/2)

EF = 4.5 cm

(c) CG = 2DF

CG = 2 x 4.2

CG = 8.4 cm

(d) AE = 2BD

BD = (1/2) AE

BD = (1/2) x 12

We get,

BD = 6 cm

14. The diagonals AC and BD of a quadrilateral ABCD intersect at right angles. Prove that the quadrilateral formed by joining the mid-points of quadrilateral ABCD is a rectangle.

Solution:

The figure is as shown below

FRANK Solutions Class 9 Maths Chapter 15 - 17

Let ABCD be a quadrilateral where P, Q, R, and S are the mid-points of sides AB, BC, CD, and DA. Diagonals AC and BD intersect at point ‘O’.

We need to prove that PQRS is a rectangle

Proof:

In △ABC and △ADC,

2PQ = AC and PQ || AC …….. (1)

2RS = AC and RS || AC …….. (2)

From (1) and (2)

We get,

PQ = RS and PQ || RS

Similarly, we can show that

PS = RQ and PS || RQ

Hence,

PQRS is a parallelogram

PQ || AC

Therefore, ∠AOD = ∠PXO = 900 …. [Corresponding angles]

Again BD || RQ

Therefore, ∠PXO = ∠RQX = 900 …[Corresponding angles]

Similarly,

∠QRS = ∠RSP = ∠SPQ = 900

Hence,

PQRS is a rectangle

15. In △ABC, D and E are the midpoints of the sides AB and AC, respectively. F is any point on the side BC. If DE intersects AF at P, show that DP = PE.

Solution:

Note: The given question is incomplete

According to the question given, F could be any point on BC, as shown below

FRANK Solutions Class 9 Maths Chapter 15 - 18

So, this makes it impossible to prove DP = DE

Since P too would shift as F shift because P too would be any point on DE as F is

Note: If we are given F to be the mid-point of BC, the result can be proved

FRANK Solutions Class 9 Maths Chapter 15 - 19

Here,

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

DE || BC and DE = (1/2) BC

But F is the mid-point of BC

BF = FC = (1/2) BC = DE

Since D is the mid-point of AB, and DP || BF

Since P is the mid-point of AF and E is the mid-point of AC,

PE = (1/2) FC

Also,

D and P are the mid-points of AB and AF, respectively

 DP = (1/2) BF = (1/2) FC = PE …….. (since BF = FC)

 DP = PE

Hence, proved.

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