**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

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

**1. In â–³ABC, D is the mid-point of AB, and E is the mid-point of BC**

**Calculate:**

**(i) DE, if AC = 8.6 cm**

**(ii) âˆ DEB, if âˆ ACB = 72 ^{0}**

**Solution:**

In â–³ABC,

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) âˆ DEB = âˆ C = 72^{0} (Corresponding angles are equal, since DE || AC)

**2. In â–³ABC, 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:**

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 â–³ABC, D, E, and F are the mid-points of BC, CA and AB, respectively. Find FE, if BC = 14 cm**

**(b) In â–³ABC, D, E, and F are the mid-points of BC, CA and AB, respectively. Find DE, if AB = 8 cm**

**(c) In â–³ABC, D, E, and F are the mid-points of BC, CA and AB, respectively. Fine âˆ FDB if âˆ ACB = 115 ^{0}**

**Solution:**

(a)

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

Hence,

FE = (1/2) BC â€¦â€¦â€¦â€¦ (Mid-point Theorem)

= (1/2) x 14

We get,

= 7 cm

(b)

In â–³ABC,

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

Hence,

DE = (1/2) AB â€¦â€¦.. (Mid-point Theorem)

= (1/2) x 8

We get,

= 4 cm

(c)

In â–³ABC,

FD || AC

Hence,

âˆ FDB = âˆ ACB = 115^{0} â€¦â€¦.(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**

**Solution:**

In â–³NSR,

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 â–³ABC, 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:**

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 â–³ACQ,

EF|| AQ and EF = (1/2) AQ â€¦â€¦â€¦(1)

In â–³ABP,

EF || AP and EF = (1/2) AP â€¦â€¦â€¦.(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

â‡’Â (1/2) AQ = (1/2) AP

**â‡’**Â 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:**

In the given rectangle ABCD,

Join AC and BD

In â–³ABC, P and Q are the mid-points of AB and BC, respectively

PQ = (1/2) AC â€¦â€¦.. (1) and PQ ||AC

In â–³BDC, R and Q are the mid-points of CD and BC, respectively

QR = (1/2) BD â€¦â€¦..(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 â–³ABC in which AB = BC. Prove that â–³DEF is also** **isosceles.**

**Solution:**

E and F are mid-points of BC and AC

Hence, EF = (1/2) AB â€¦â€¦.(1)

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

Hence, DF = (1/2) BC â€¦â€¦..(2)

But AB = BC

From (1) and (2)

EF = DF

Thus, â–³DEF 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:**

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

Hence,

PQ || AC and PQ = (1/2) AC â€¦â€¦. (i)

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

Hence,

SR || AC and SR = (1/2) AC â€¦â€¦.(ii)

From (i) and (ii),

PQ || SR and PQ = SR

Therefore, PQRS is a parallelogram

Further, AC and BC intersect at right angles

âˆ´ SP || BD and BD âŠ¥ AC

âˆ´ SP âŠ¥ AC

â‡’Â SP âŠ¥ SR

â‡’Â âˆ RSP = 90^{0}

âˆ´ âˆ RSP = âˆ SRQ = âˆ RQP = âˆ SPQ = 90^{0}

Hence, PQRS is a rectangle

**9. In a right-angled triangle ABC. âˆ ABC = 90 ^{0,} and D is the mid-point of AC. Prove that BD = (1/2) AC**

**Solution:**

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

Now,

DE || CB and AB are the transversal,

âˆ´âˆ AED = âˆ ABC â€¦. (corresponding angles)

âˆ ABC = 90^{0} (given)

â‡’Â âˆ AED = 90^{0}

Also,

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

DE bisects side AB,

Hence,

AE = BE â€¦..(i)

In â–³AED and â–³BED,

âˆ AED = âˆ BED â€¦..(Each 90^{0})

AE = BE â€¦â€¦[From (i)]

DE = DE â€¦.(Common)

Therefore, â–³AED **â‰… **â–³BED â€¦.(By SAS)

â‡’Â AD = BD â€¦..(C.P.C.T.C)

â‡’Â 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) â–³GEA â‰… â–³GFD**

**(b) â–³HEB â‰… â–³HFC**

**(c) EGFH is a parallelogram**

**Solution:**

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 â€¦â€¦â€¦.(1)

(a) In â–³GEA and â–³GFD,

AE = DF â€¦â€¦ [From (1)]

âˆ AGE = âˆ DGF â€¦â€¦â€¦ (vertically opposite angles)

âˆ GAE = âˆ GFD â€¦â€¦.. (Alternate interior angles)

Therefore,

â–³GEA â‰… â–³GFD

(b) In â–³HEB and â–³HFC,

BE = FC â€¦..[From (1)]

âˆ EHB = âˆ FHC (vertically opposite angles)

âˆ HBE = âˆ HFC (Alternate interior angles)

Therefore,

â–³HEB â‰… â–³HFC

(c) In quadrilateral AECF,

AE = CF â€¦..[From (1)]

AE || CF â€¦…(since AB || DC)

Hence,

AECF is a parallelogram

**â‡’**Â EC || AF or EH || GF â€¦â€¦ (i)

In quadrilateral BFDE,

BE = DF â€¦.[From (1)]

BE || DF â€¦.(since AB || DC)

â‡’ BEDF is a parallelogram

**â‡’**Â BF || ED or HF || EG â€¦â€¦(ii)

From equations (i) and (ii),

We get,

EGFH is a parallelogram

**11. In â–³ABC, 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:**

In â–³BDC and â–³ADQ,

CD = DQ â€¦ (given)

âˆ BDC = âˆ ADQ â€¦â€¦(vertically opposite angles)

BD = AD â€¦â€¦(D is the mid-point of AB)

Therefore,

â–³BDC â‰… â–³ADQ

**â‡’**Â âˆ DBC = âˆ DAQ (cpct)â€¦â€¦.(i)

And, BC = AQ (cpct) â€¦â€¦.(ii)

Similarly,

We can prove

â–³CEB â‰… â–³ AEP

**â‡’**Â âˆ ECB = âˆ EAP (cpct)â€¦â€¦..(iii)

And, BC = AP (cpct)â€¦â€¦..(iv)

(a) In â–³ABC,

âˆ ABC + âˆ ACB + âˆ BAC = 180^{0}

**â‡’**Â âˆ DBC + âˆ ECB + âˆ BAC = 180^{0}

**â‡’**Â âˆ DAQ + âˆ EAP + âˆ BAC = 180^{0} [From (i) and (iii)]

**â‡’**Â Q, A, P are collinear

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

AQ = AP

Therefore, A is the mid-point of PQ

**12. In â–³ABC, 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**

**Solution:**

In â–³AEG,

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

Hence,

F is the mid-point of AG

AF = FG â€¦â€¦â€¦.. (1)

In â–³ABC,

DF || EG|| BC

DE = BE

Hence,

GF = GC â€¦â€¦â€¦â€¦. (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**

**Solution:**

According to the equal intercept theorem,

Since CD = DE

AB = BC â€¦â€¦â€¦(i)

EF = GF â€¦â€¦â€¦.(ii)

(a) BC = AB = 6 cm â€¦.. [From (i)]

(b) EG = EF + FG

EG = 2EF â€¦… [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

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 â€˜Oâ€™.

We need to prove that PQRS is a rectangle

Proof:

In â–³ABC and â–³ADC,

2PQ = AC and PQ || AC â€¦â€¦.. (1)

2RS = AC and RS || AC â€¦â€¦.. (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, âˆ AOD = âˆ PXO = 90^{0} â€¦. [Corresponding angles]

Again BD || RQ

Therefore, âˆ PXO = âˆ RQX = 90^{0} â€¦[Corresponding angles]

Similarly,

âˆ QRS = âˆ RSP = âˆ SPQ = 90^{0}

Hence,

PQRS is a rectangle

**15. In â–³ABC, 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

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

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

**â‡’**Â DP = (1/2) BF = (1/2) FC = PE â€¦â€¦.. (since BF = FC)

**â‡’**Â DP = PE

Hence, proved.

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