ICSE Solutions for Selina Concise Chapter 12 Mid Point and its Converse (including Intercept Theorem) Class 9 Maths

Exercise 12(A)

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,

Since, M is the midpoint of AB and MN ॥ BC hence N is the midpoint of AC.
Therefore,
MN = (1/2)BC = (1/2)×7 = 3.5cm
and  AN = (1/2)AC = (1/2)×5 = 2.5 cm

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,

Let ABCD be a rectangle where P,Q,R,S are the midpoint of AB, BC, CD, DA. We need to show that PQRS is a rhombus for help we draw two diagonal BD and AC as shown in figure .
where BD = AC (Since diagonal of rectangle are equal)
Proof :
From △ABD and △BCD
PS = (1/2)BD = QR and PS ॥ BD ॥ QR
2PS = 2QR = BD and PS ॥ QR ...(1)

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

3. D, E and F are the mid-points of the sides AB, BC and CA of an isosceles Î”ABC in which AB = BC. Prove that Î”DEF 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

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) we get,

2(PQ+QR) = AB+CD

⇒ PR = (1/2)(AB + CD)

Hence, proved

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 us draw a diagonal AC as shown in the figure below,

(i) Given that,
AB = 11 cm, CD = 8 cm

From triangle ABC,
ON = (1/2)AB = (1/2)×11 = 5.5cm

From triangle ACD,
Om = (1/2)CD = (1/2)×8 = 4 cm
Hence MN = OM + ON = (4+5.5) = 9.5cm

(ii) Given that,
CD = 20 cm , MN = 27 cm

From triangle ACD,
OM = (1/2)CD = (1/2)×20 = 10 cm
Therefore ON = 27-10 = 17cm

From triangle ABC,
AB = 2 × ON = 2× 17 = 34 cm

(iii) Given that,
AB = 23cm, MN = 15cm

From triangle ABC,
ON = (1/2)AB = (1/2)×23 = 11.5 cm
Therefore, OM = 15 - 11.5 = 3.5cm
From triangle ACD,
CD = 2×OM  = 2×3.5 = 7cm

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

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
Therefore PQRS is a parallelogram.
Now PQ॥ AC, therefore ∠AOD = ∠PXO = 90° [Corresponding angle]
Again BD ॥ RQ, therefore ∠PXO = ∠RQX = 90° [Corresponding angle]
Similarly, ∠QRS = ∠RSP  = ∠SPQ = 90°

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

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

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) PR = (1/2)DB

For help we draw the diagonal BD as shown below

The diagonal AC and BD cuts at point x.
We know that the diagonal of a parallelogram intersects equally each other. Therefore
AX = CX and BX = DX
Given,
CQ = (1/4)AC
⇒ CQ = (1/4)× 2CX
⇒ CQ = (1/2)CX
Therefore Q is the midpoint of CX.
(i) For triangle CDX  PQ॥ DX  or PR॥ BD.
Since, for triangle CBX Q is the midpoint of CX and QR॥BX.
Therefore R is the midpoint of BC
(ii) For triangle BCD
As P and R are the midpoint of CD and BC, therefore PR= (1/2)DB

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

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

From the figure it is seen that P is the midpoint of BC and PQ ॥ AC and QR ॥ BC.
Therefore Q is the midpoint of AB and R is the midpoint of AP
(i) Therefore AP = 2AR
(ii) Here we increase QR so that it cuts AC at S as shown in the figure.
(iii) From triangle PQR and triangle ARS
∠PQR = ∠ARS (Opposite angle )
PR = AR
PQ = AS  [PQ = AS = (1/2)AC]
△PQR ≅ △ARS  (SAS Postulate)
Therefore QR = RS
Now
BC = 2QS
⇒ BC = 2×2QR
⇒ BC = 4QR
Hence, proved

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

(i) From △PED and △ABP
PD = AP  [P is the midpoint of AD]
∠DPE = ∠APB  [Opposite angle]
∠PED = ∠PBA  [AB ॥ CE]
∴ Î”PED ≅ Î”ABP  [ASA postulate]
∴ EP = BP

(ii) For triangle ECB PQ||CE

Again, CE||AB

Therefore, PQ||AB

Hence, proved

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, i.e. AF = GF ...(1)

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

Therefore G is the midpoint of CF, i.e. GF = CF …(2)

AC = AF + GF + CF

⇒ AC = 3AF (From 1 and 2)

Hence, proved

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

(i) Since F is the midpoint and EF ॥ AB.
Therefore E is the midpoint of BC
So,
BE = (1/2)BC and EF = (1/2)AB  ...(1)
Since, D and F are the midpoint of AB and AC
Therefore,
DE ॥ BC
So DF = (1/2)BC and DB = (1/2)AB  ...(2)
From (1), (2) we get
BE = DF and BD = EF
Hence, BDEF is a parallelogram.

(ii) Since
AB = 2EF
= 2×4.8
= 9.6cm

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.

In Î”ABC,
AD is the median of BC
⇒ D is the mid-point of BC.
Given that DE P BA
By the Converse of the Mid - point theorem,
⇒ DE bisects AC
⇒ E is the mid- point of AC
⇒ BE is the median of AC, that is BE is also a median.

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.

Construction: Draw DY ॥ BQ
In Î”BCQ and Î”DCY,
ã„¥BCQ = ã„¥DCY  (Common)
ã„¥BQC = ã„¥DYC  (Corresponding angles)
So, Î”BCQ ～Î”DCY  (AA Similarity criterion)
⇒ BQ/DY = BC/DC = CQ/CY  (Corresponding sides are proportional)
⇒ BQ/DY = 2CD/CD  (D is the mid - point of BC)
⇒ BQ/DY = 2  ...(i)
⇒ EQ/DY = AE/ED = 1/2  (E is the mid-point of AD)
that is EQ/DY = 1/2  ...(ii)
Dividing (i) by (ii), we get
⇒ BQ/EQ = 4
⇒ BE + EQ = 4EQ
⇒ BE = 3EQ
⇒ BE/EQ = 3/1

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.

In Î”EDF,
Mis the mid- point of AB and Nis the mid- Point of DE.
⇒ MN = (1/2)EF (Mid-point theorem)
⇒ EF = 2MN ...(i)
In Î”ABC,
M is the mid - point of AB and N is the mid- point of BC
⇒ MN = (1/2)AC  (Mid - point theorem)
⇒ AC = 2MN  ...(ii)
From (i) and (ii), we get
⇒ EF = AC

Exercise 12(B)

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

(iii) AE = 28D = 2×4.1 = 8.2
(iv) DF = (1/2) CG = (1/2)×11 = 5.5cm

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

From △BRS and △QRC
∠BRS = ∠QRC
BR = RC
∠RBS = ∠RCQ
∴ △BRS ≅ △QRC
∴ QR = RS
⇒ DS = DQ + QR + RS = QR + QR + RS = 3RS

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:

Here, D is the midpoint of BC and DP is parallel to AB, therefore P is the midpoint of AC and PD = (1/2)AB
(i) Again from the triangle AEF we have AE ॥ PD ॥ CR and AP = (1/3)AE
Therefore DF = (1/3)EF or we can say that 3DF = EF.
Hence it is shown.
(ii) From the triangle PED we have PD ॥ CR and C is the midpoint of PE therefore CR = (1/2)PD
Now,
PD = (1/2)AB
⇒ (1/2)PD = (1/4)AB
⇒ CR = (1/4)AB
⇒ 4CR = AB
Hence, it is shown.

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

(i) From triangle BPC and triangle APN
∠BPC = ∠APN  [Opposite angle]
BP = AP
PC = PN
∴ Î” BPC ≅ Î”APN [SAS postulate]
∴ ∠PBC =∠PAN  ...(1)
And BC = AN  ...(3)
Similarly ∠QCB = ∠QAN ...(2)
And BC = AM  ...(4)
Now
∠ABC + ∠ACB + ∠BAC = 180°
∠PAN + ∠QAM + ∠BAC = 180°  [(1), (2) we get]
Therefore M, A, N are collinear
(ii) From (3) and (4) MA = NA
Hence, A is the midpoint of MN

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.

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

(i) From △HEB and △FHC
BE = FC
∠EHB = ∠FHC  [Opposite angle]
∠HBE = ∠HFC
∴ △HEB ≅ △FHC
∴ EH = CH, BH = FH
(ii) Similarly AG = GF and EG = DG  ...(1)
For triangle ECD, F and H are the midpoint of CD and EC.
Therefore HF॥DE and HF = (1/2) DE  ...(2)

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

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

Therefore, GEHF is a parallelogram.

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.

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

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

Let ABCD be a quadrilateral where P,Q, R, S ae the midpoint of AB, BC, CD, DA, PQRS is a rectangle.
Diagonal AC and BD intersect at point O. We need to show that AC and BD intersect at right angle.
Proof :
PQ ॥ AC, therefore ∠AOD = ∠PXO [Corresponding angle] ...(1)
Again BD॥RQ, therefore ∠PXO = ∠RQX = 90° [corresponding angle and angle of rectangle]...(2)
From (1) and (2) we get,
∠AOD = 90°
Similarly ∠AOB = ∠BOC = ∠DOC = 90°
Therefore, diagonals AC and BD intersect at right angle
Hence, proved

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

From figure,
Since E is the midpoint of AC and EF॥ AB
Therefore F is the midpoint of BC and 2 DE = BC or DE = BF
Again,
D and E are midpoint, therefore DE॥ BF and EF॥ BD
Hence BDEF is a parallelogram.
Now,
BD = EF = (1/2)AB = (1/2)×16 = 8cm
BF = DE = (1/2)BC = (1/2)×18 = 9cm
Therefore, perimeter of BDEF = 2 (BF + EF) = 2(9+8) = 34 cm

11. In the given figure, AD and CE are medians and DF||CE. Prove that: FB = (1/4)AB.

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.

So, FB = (1/2)BE but BE = (1/2)AB
Substitute value of BE in first equation, we get
FB = (1/4)AB
Hence, Proved

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

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

14. In Î”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 Î”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 Î” ABC.

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.