# Frank Solutions for Chapter 15 Mid-point and Intercept Theorems Class 9 Mathematics ICSE

### Exercise 15.1

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

Calculate:

(i) DE, if AC = 8.6 cm

(ii) DEB, if ACB = 72°

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) ×8.6

We get,

= 4.3 cm

(ii) ∠DEB = ∠C = 72° (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?

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, F are the mid-points of BC, CA and AB respectively. Find FE, if BC = 14 cm

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

(c) In ABC, D, E, F are the mid-points of BC, CA and AB respectively. Fine FDB if ACB = 115°

(a)

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)

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) ×8

We get,

= 4 cm

(c)

In △ABC,

FD || AC

Hence,

∠FDB = ∠ACB = 115° …(Corresponding angles are equal)

4. In parallelogram PQRS, L is 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.

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

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.

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.

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.

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 = 90°

∴ ∠RSP = ∠SRQ = ∠RQP = ∠SPQ = 90°

Hence, PQRS is a rectangle

9. In a right angled triangle ABC. ABC = 90° and D is the mid-point of AC. Prove that BD = (1/2) AC

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

Now,

DE || CB and AB is the transversal,

∴ ∠AED = ∠ABC …(corresponding angles)

∠ABC = 90° (given)

⇒ ∠AED = 90°

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 90°)

AE = BE …[From (i)]

DE = DE …(Common)

Therefore, △AED △BED …(By SAS)

⇒ 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

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

AE = CF …[From (1)]

AE || CF …(since AB || DC)

Hence,

AECF is a parallelogram

EC || AF or EH || GF …(i)

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

CD = DQ …(given)

∠BDC = ∠ADQ …(vertically opposite angles)

BD = AD …(D is the mid-point of AB)

Therefore,

∠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 = 180°

∠DBC + ∠ECB + ∠BAC = 180°

∠DAQ + ∠EAP + ∠BAC = 180° [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 meet the side BC at points M and N respectively. Prove that BM = MN = NC

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

According to 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 = 2×4.2

⇒ CG = 8.4 cm

(d) AE = 2BD

⇒ BD = (1/2) AE

⇒ BD = (1/2)×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.

The figure is as shown below

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

We need to prove that PQRS is a rectangle

Proof:

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 = 90° …[Corresponding angles]

Again BD || RQ

Therefore, ∠PXO = ∠RQX = 90° …[Corresponding angles]

Similarly,

∠QRS = ∠RSP = ∠SPQ = 90°

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.

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

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

DP = PE

Hence, proved

### Exercise 15.2

1. Prove that the straight lines joining the mid-points of the opposite sides of a quadrilateral bisect each other.

2. If L and M are the mid-points of AB, and DC respectively of parallelogram ABCD. Prove that segments DL and BM trisect diagonal AC.

3. In a right-angled triangle ABC. ABC = 90° and D is the midpoint of AC. Prove that BD = 1/2.AC.

4. In parallelogram ABCD, P is the mid-point of DC. Q is point on AC such that CQ = 1/4 AC. PQ produced meets BC at R. Prove that

(i) R is the mid-point of BC, and

(ii) PR = ½.DB.

5. In a parallelogram ABCD, M is the mid-point of AC. X and Y are the points on AB and DC respectively such that AX = CY. Prove that:

(i) Triangle AXM is congruent to triangle CYM, and

(ii) XMY is a straight line.

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

7. In the given figure, ABCD is a trapezium. P and Q are the midpoints of non-parallel side AD and BC respectively. Find :

(a) PQ, if AB = 12 cm and DC = 10 cm.

(b) AB, if DC = 8 cm and PQ = 9.5 cm

(c) DC, if AB = 20 cm and PQ = 14 cm

(a)

(b)

(c)

8. In AD is a median of side BC of ABC. E is the midpoint of AD. BE is joined and produced to meet AC at F. Prove that AF : AC = 1 : 3.

Construction : Draw DS || BF, meeting AC at S.

9. M and N divide the side AB of ABC into three equal parts. Line segments MP and NQ are both parallel to BC, and meet AC at P and Q respectively. Prove that P and Q divide AC into three equal parts.

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

11. ABCD is a parallelogram. E is the mid-point of CD and P is a point on AC such that PC = 1/4AC. EP produced meets at BC at F. Prove that:

(i) F is the mid-point of BC

(ii) ZEF = BD

(i)

(ii)

12. ABCD is a kite in which BC = CD, AB = AD. E, F and G are the mid-points of CD, BC and AB respectively. Prove that:

(i) EFG = 90°

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

(i)

(ii)

13. In ABC, X is the mid-point of AB, and Y is the mid-point of AC, BY and CX are produced and meet the straight line through A parallel to BC at P and Q respectively. Prove AP = AQ.

14. In ABC, D, E and F are the midpoints of AB, BC and AC.

(a) Show that AE and DF bisect each other.

(b) If AE and DF intersect at G, and M and N are the midpoints of GB and GC respectively, prove that DMNF is a parallelogram.