Selina Concise Class-9th Mid Point and Intercept Theorem ICSE Mathematics Solutions Chapter-12. We provide step by step Solutions of Exercise / lesson-12 Mid Point and its Converse ( Including Intercept Theorem ) for ICSE Class-9 Concise Selina Mathematics by R K Bansal.

Our Solutions contain all type Questions with Exe-12 A and Exe-12 B, to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board Class-9 Mathematics .

## Selina Concise Class-9th Mid Point and Intercept Theorem ICSE Mathematics Solutions Chapter-12

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Exe – 12 A,

Exe -12 B,

### Exercise – 12 A, Selina Concise Class-9th Mid Point and its Converse ( Including Intercept Theorem ) ICSE Mathematics Solutions

#### Question 1

In triangle ABC, M is mid-point of AB and a straight line through M and parallel to BC cuts AC in N. Find the lengths of AN and MN if Bc = 7 cm and Ac = 5 cm.

The triangle is shown below,

#### Question 2

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

The figure is shown below,

Similarly 2PQ=2SR=AC and PQ||SR—– (2)

From (1) and (2) we get

PQ=QR=RS=PS

Therefore PQRS is a rhombus.

Hence proved

#### Question 3

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

The figure is shown below

Given that ABC is an isosceles triangle where AB=AC.

Since D,E,F are midpoint of AB,BC,CA therefore

2DE=AC and 2EF=AB this means DE=EF

Therefore DEF is an isosceles triangle an DE=EF.

Hence proved

#### Question 4

The following figure shows a trapezium ABCD in which AB // DC. P is the mid-point of AD and PR // AB. Prove that:

…………………

Here from triangle ABD P is the midpoint of AD and PR||AB, therefore Q is the midpoint of BD

Similarly R is the midpoint of BC as PR||CD||AB

From triangle ABD 2PQ=AB …….(1)

From triangle BCD 2QR=CD …..(2)

Now (1)+(2)=>

2(PQ+QR)=AB+CD

Hence proved

#### Question 5

The figure, given below, shows a trapezium ABCD. M and N are the mid-point of the non-parallel sides AD and BC respectively. Find:

……………

(i) MN, if AB = 11 cm and DC = 8 cm.

(ii) AB, if Dc = 20 cm and MN = 27 cm.

(iii) DC, if MN = 15 cm and AB = 23 cm.

Let we draw a diagonal AC as shown in the figure below,

#### Question 6

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

The figure is shown below

#### Question 7

L and M are the mid-point of sides AB and DC respectively of parallelogram ABCD. Prove that segments DL and BM trisect diagonal AC.

The required figure is shown below

From figure,

BL=DM and BL||DM and BLMD is a parallelogram, therefore BM||DL

From triangle ABY

L is the midpoint of AB and XL||BY, therefore x is the midpoint of AY.ie AX=XY …..(1)

Similarly for triangle CDX

CY=XY …..(2)

From (1) and (2)

AX=XY=CY and AC=AX+XY+CY

Hence proved

#### Question 8

ABCD is a quadrilateral in which AD = BC. E, F, G and H are the mid-points of AB, BD, CD and Ac respectively. Prove that EFGH is a rhombus.

……………………..

From the figure,

For triangle ADC and triangle ABD

For triangle BCD and triangle ABC

2GF=BC and 2EH=BC, therefore 2GF=2EH=BC …..(3)

From (1),(2),(3) we get,

2GH=2EF=2GF=2EH

GH=EF=GF=EH

Therefore EFGH is a rhombus.

Hence proved

#### Question 9

A parallelogram ABCD has P the mid-point of Dc and Q a point of Ac such that CQ = 1/4 AC. PQ produced meets BC at R.

Prove that

(i)R is the midpoint of BC

(ii)……….

For help we draw the diagonal BD as shown below

#### Question 10

D, E and F are the mid-points of the sides AB, BC and CA respectively of ABC. AE meets DF at O. P and Q are the mid-points of OB and OC respectively. Prove that DPQF is a parallelogram.

The required figure is shown below

For triangle ABC and OBC

2DE=BC and 2PQ=BC, therefore DE=PQ …..(1)

For triangle ABO and ACO

2PD=AO and 2FQ=AO, therefore PD=FQ …..(2)

From (1),(2) we get that PQFD is a parallelogram.

Hence proved

#### Question 11

In triangle ABC, P is the mid-point of side BC. A line through P and parallel to CA meets AB at point Q; and a line through Q and parallel to BC meets median AP at point R. Prove that

(i)AP=2AR

(ii)BC=4QR

The required figure is shown below

#### Question 12

In trapezium ABCD, AB is parallel to DC; P and Q are the mid-points of AD and BC respectively. BP produced meets CD produced at point E. Prove that:

(i) Point P bisects BE,

(ii) PQ is parallel to AB.

The required figure is shown below

(ii)For tiangle ECB PQ||CE

Again CE||AB

Therefore PQ||AB

Hence proved

#### Question 13

In a triangle ABC, AD is a median and E is mid-point of median AD. A line through B and E meets AC at point F.

Prove that: AC = 3AF

The required figure is shown below

For help we draw a line DG||BF

Now from triangle ADG, DG||BF and E is the midpoint of AD

Therefore F is the midpoint of AG,ie AF=GF …..(1)

From triangle BCF, DG||BF and D is the midpoint of BC

Therefore G is the midpoint of CF,ie GF=CF …(2)

AC=AF+GF+CF

AC=3AF(From (1) and (2))

Hence proved

#### Question 14

D and F are mid-points of sides AB and AC of a triangle ABC. A line through F and parallel to AB meets BC at point E.

(i) Prove that BDFE is parallelogram

(ii) Find AB, if EF = 4.8 cm.

The required figure is shown below

#### Question 15

In triangle ABC, AD is the median and DE, drawn parallel to side BA, meets AC at point E. Show that BE is also a median.

#### Question 16

In ∆ABC, E is mid-point of the median AD and BE produced meets side AC at point Q. Show that BE : EQ = 3:1.

#### Question 17

In the given figure, M is mid-point of AB and DE, whereas N is mid-point of BC and DF. Show that: EF = AC.

### Selina Concise Class-9th Mid Point and Intercept Theorem ICSE Mathematics Solutions Exercise – 12 B

#### Question 1

Use the following figure to find:

(i) BC, if AB = 7.2 cm.

(ii) GE, if FE = 4 cm.

(iii) AE, if BD = 4.1 cm

(iv) DF, if CG = 11 cm.

According to equal intercept theorem since CD=DE

Therefore AB=BC and EF=GF

(i)BC=AB=7.2cm

(ii)GE=EF+GF=2EF==8cm

Since B,D,F are the midpoint and AE||BF||CG

Therefore AE=2BD and CG=2DF

#### Question 2

In the figure, give below, 2AD = AB, P is mid-point of AB, Q is mid-point of DR and PR // BS. Prove that:

(i) AQ // BS

(ii) DS = 3 Rs.

Given that AD=AP=PB as 2AD=AB and p is the midpoint of AB

(i)From triangle DPR, A and Q are the midpoint of DP and DR.

Therefore AQ||PR

Since PR||BS ,hence AQ||BS

(ii)From triangle ABC, P is the midpoint and PR||BS

Therefore R is the midpoint of BC

#### Question 3

The side AC of a triangle ABC is produced to point E so that CE = 1/2AC. D is the mid-point of BC and ED produced meets AB at F. Lines through D and C are drawn parallel to AB which meet AC at point P and EF at point R respectively. Prove that:

(i) 3DF = EF(ii) 4CR = AB.

Consider the figure:

#### Question 4

In triangle ABC, the medians BP and CQ are produced upto points M and N respectively such that BP = PM and CQ = QN. Prove that:

(i) M, A and N are collinear.

(ii) A is the mid-point of MN

The figure is shown below

#### Question 5

In triangle ABC, angle B is obtuse. D and E are mid-points of sides AB and BC respectively and F is a point on side AC such that EF is parallel to AB. Show that BEFD is a parallelogram.

The figure is shown below

From the figure EF||AB and E is the midpoint of BC.

Therefore F is the midpoint of AC.

Here EF||BD, EF=BD as D is the midpoint of AB

BE||DF, BE=DF as E is the midpoint of BC.

Therefore BEFD is a parallelogram.

#### Question 6

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

(i) Triangles HEB and FHC are congruent;

(ii) GEHF is a parallelogram.

The figure is shown below

(1),(2) we get, HF=EG and HF||EG

Similarly we can show that EH=GF and EH||GF

Therefore GEHF is a parallelogram.

#### Question 7

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

The figure is shown below

For triangle AEG

D is the midpoint of AE and DF||EG||BC

Therefore F is the midpoint of AG.

AF=GF …..(1)

Again DF||EG||BC DE=BE, therefore GF=GC …..(2)

(1),(2) we get AF=GF=GC.

Similarly Since GN||FM||AB and AF=GF ,therefore BM=MN=NC

Hence proved

#### Question 8

In triangle ABC; M is mid-point of AB, N is mid-point of AC and D is any point in base BC. Use intercept Theorem to show that MN bisects AD.

The figure is shown below

Since M and N are the midpoint of AB and AC, MN||BC

According to intercept theorem Since MN||BC and AM=BM,

Therefore AX=DX. Hence proved

#### Question 9

If the quadrilateral formed by joining the mid-points of the adjacent sides of quadrilateral ABCD is a rectangle, show that the diagonals AC and BD intersect at right angle.

The figure is shown below

#### Question 10

In triangle 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 = 16 cm, AC = 12 cm and BC = 18 cm, find the perimeter of the parallelogram BDEF.

The figure is shown below

#### Question 11

In the given figure, AD and CE are medians and DF//CE. Prove that: ………

………………

Given AD and CE are medians and DF || CE.

We know that from the midpoint theorem, if two lines are parallel and the starting point of segment is at the midpoint on one side, then the other point meets at the midpoint of the other side.

Consider triangle BEC. Given DF || CE and D is midpoint of BC. So F must be the midpoint of BE.

#### Question 12

In parallelogram ABCD, E is the mid-point of AB and AP is parallel to EC which meets DC at point O and BC produced at P.

Prove that:

(ii) O is the mid-point of AP.

………………

Given ABCD is parallelogram, so AD = BC, AB = CD.

Consider triangle APB, given EC is parallel to AP and E is midpoint of side AB. So by midpoint theorem, C has to be the midpoint of BP.

So BP = 2BC, but BC = AD as ABCD is a parallelogram.

Consider triangle APB, AB || OC as ABCD is a parallelogram. So by midpoint theorem, O has to be the midpoint of AP.

Hence Proved

#### Question 13

In trapezium ABCD, sides AB and DC are parallel to each other. E is mid-point of AD and F is mid-point of BC. Prove that: AB + DC = 2EF.

Consider trapezium ABCD.

Given E and F are midpoints on sides AD and BC, respectively.

We know that AB = GH = IJ

From midpoint theorem,

Consider LHS,

AB + CD = AB + CJ + JI + ID = AB + 2HF + AB + 2EG

So AB + CD = 2(AB + HF + EG) = 2(EG + GH + HF) = 2EF

AB + CD = 2EF

Hence Proved

#### Question 14

In Δ ABC, AD is the median and DE is parallel to BA, where E is a point in AC. Prove that BE is also a median.

Given Δ ABC

AD is the median. So D is the midpoint of side BC.

Given DE || AB. By the midpoint theorem, E has to be midpoint of AC.

So line joining the vertex and midpoint of the opposite side is always known as median. So BE is also median of Δ ABC.

#### Question 15

Adjacent sides of a parallelogram are equal and one of the diagonals is equal to any one of the sides of this parallelogram. Show that its diagonals are in the ratio.

— End of Mid Point and its Converse ( Including Intercept Theorem ) Solutions :–

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