SELINA Solution Class 9 Chapter 12 Mid-point and its Converse [Including intercept Theorem]Exercise 12A

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.

Sol:

The triangle is shown below,

Since M is the midpoint of AB and MN || BC hence N is the midpoint of AC. Therefore

MN = ${\displaystyle \frac{1}{2}}$ BC = ${\displaystyle \frac{1}{2}}$ x 7 = 3.5cm

And AN = ${\displaystyle \frac{1}{2}}$ AC = ${\displaystyle \frac{1}{2}}$  x 5 = 2.5cm

Question 2

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

Sol:

Given: Let ABCD be a rectangle where P, Q, R, S are the midpoint of AB, BC, CD, DA.

To Prove: PQRS is a rhombus

Construction: Draw two diagonal BD and AC as shown in figure. Where BD = AC ( Since diagonal of the rectangle are equal )

Proof:
From ΔABD and ΔBCD 
PS = ${\displaystyle \frac{1}{2}}$ 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.

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.

Sol:

The figure is shown below

Given that ABC is an isosceles triangle where AB = AC.
Since D, E, F are the mid-point of AB, BC, CA therefore
2DE = AC and 2EF = AB this means DE = EF
Therefore DEF is an isosceles triangle a 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:

PR = ${\displaystyle \frac{1}{2}}$ 9 ( AB + CD)

Sol:

Here from the 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
PR = ${\displaystyle \frac{1}{2}}$ ( 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.

Sol:

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

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

ON = ${\displaystyle \frac{1}{2}}$ AB =${\displaystyle \frac{1}{2}}$ x 11 = 5.5 cm

From triangle ACD

OM = ${\displaystyle \frac{1}{2}}$ CD =${\displaystyle \frac{1}{2}}$ x 8 = 4 cm

Hence MN = OM + ON = ( 4 + 5.5 ) = 9.5 cm

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

From triangle ACD

OM =${\displaystyle \frac{1}{2}}$ CD =${\displaystyle \frac{1}{2}}$ x 20 =10 cm

Therefore ON = 27 - 10 = 17 cm

From triangle ABC

AB = 2ON  = 2 x 17 = 34 cm

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

From triangle ABC

ON =${\displaystyle \frac{1}{2}}$ AB =${\displaystyle \frac{1}{2}}$ x 23 = 11.5 cm

Therefore OM = 15 - 11.5 = 3.5 cm

From triangle ACD

CD = 2O M = 2 x 3.5 = 7 cm

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.

Sol:

The figure is shown below

Let ABCD be a quadrilateral where P, Q, R, S are the midpoint of AB, BC, CD, DA. Diagonal AC and BD intersect at a right angle at point O. We need to show that PQRS is a rectangle

Proof:

From and ΔABC and ΔADC
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 angel ]

Again BD || RQ, therefore ∠PXO = ∠RQX = 90°  ...[ Corresponding angel]

Similarly ∠QRS = ∠RSP = ∠SPQ = 90°  
Therefore PQRS is a rectangle.
Hence proved.

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.

Sol:

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.

Sol:

Given that AD = BC                                                    …..(1)

From the figure,
For triangle ADC and triangle ABD

2GH = AD and 2EF = AD, therefore 2GH = 2EF = AD …..(2)

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 = ${\displaystyle \frac{1}{4}}$AC. PQ produced meets BC at R.

Prove that
(i)R is the midpoint of BC
(ii) PR = ${\displaystyle \frac{1}{2}}$ DB

SOl:

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 intersect equally with each other. Therefore

AX = CX and BX = DX

Given,
CQ = ${\displaystyle \frac{1}{4}}$AC

CQ = ${\displaystyle \frac{1}{4}}$ x 2CX

CQ = ${\displaystyle \frac{1}{2}}$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 mid-point of CD and BC, therefore  PR = ${\displaystyle \frac{1}{2}}$ DB

Question 10

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

SOl:

Given: △ABC, D, E, F are midpoints of AB, BC, AC respectively. AB and DF meet at O. P and Q are midpoints of OB and OC respectively.
To Prove: DPFQ is a parallelogram.

Proof:
In △ABC,
D is the mid-point of AB and F is the mid-point of AC
Hence, DF ∥ BC and DF = ${\displaystyle \frac{1}{2}}$BC        ... (1) (Mid-point theorem)
In △OBC,
P is the mid-point of OB and Q is the mid-point of OC
Hence, PQ ∥ BC and PQ = ${\displaystyle \frac{1}{2}}$​BC       ... (2) (mid-point theorem)
thus, from (1) and (2)
DF ∥ PQ and DF = PQ                      ....(3)

Now, In △AOB,
D is the mid-point of AB and P is the mid-point of OB
Thus, DP ∥ AE and DP = ${\displaystyle \frac{1}{2}}$​AE        ....(4) (midpoint theorem)
 Now, In △AOC,
F is the midpoint of AC and Q is the midpoint of OC
Thus, FQ ∥ AE and QF = ${\displaystyle \frac{1}{2}}$​AE         .....(5) (midpoint theorem)
thus, from (4) and (5)
DP ∥ FQ and DP = FQ                      .....(6)

DPFQ is a parallelogram                 ......(from (3) and (6))
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

SOl:

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 = ${\displaystyle \frac{1}{2}}$AC ]
ΔPQR ≅ ΔARS                   ...( SAS Postulate )
Therefore QR = RS
Now,
BC = 2QS
BC = 2 x 2QR
BC = 4QR 
Hence proved.

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.

Sol:

The required figure is shown below

(i) From ΔPED and ΔABP,
PD = AP               ...[ P is the mid-point of AD ]
∠DPE = ∠APB      ....[ Opposite angle ]
∠PED = ∠PBA      ...[ AB || CE ]
∴ ΔPED ≅ ΔABP   ...[ ASA postulate ]
∴ EP = BP

(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.

Sol:

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.

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

SOl:

The required figure is shown below

(i) Since F is the midpoint and EF || AB.
Therefore E is the midpoint of BC.
So, ${\displaystyle \text{BE}=\frac{1}{2}\text{BC}\text{and}\text{EF}=\frac{1}{2}\text{AB}}$   …..(1)

Since D and F are the mid-point of AB and AC
Therefore DE || BC.
So, ${\displaystyle \text{DF}=\frac{1}{2}\text{BC}\text{and}\text{DB}=\frac{1}{2}\text{AB}}$  …..(2)

From (1), (2) we get
BE = DF and BD = EF
Hence  BDEF is a parallelogram.

(ii) Since
AB = 2EF
      = 2 x 4.8
      = 9.6 cm.

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.

Sol:

ln ΔABC,
AD is the median of BC.
⇒ D is the mid-point of BC.
Given at DE || 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.

Question 16

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

SOl:

Construction: Draw DY || BQ
In ΔBCQ and ΔDCY,
∠BCQ = ∠DCY                ...( Common )
∠BQC = ∠DYC                ...( Corresponding angles )
So, ΔBCQ ∼ ΔDCY          ....( AA Similarity criterion )

⇒ ${\displaystyle \frac{\text{BQ}}{\text{DY}}=\frac{\text{BC}}{\text{DC}}=\frac{\text{CQ}}{\text{CY}}}$    ..(Corresponding sides are proportional. )

⇒ ${\displaystyle \frac{\text{BQ}}{\text{DY}}=\frac{\text{2CD}}{\text{CD}}}$   ...( D is the mid-point of BC )    

⇒ ${\displaystyle \frac{\text{BQ}}{\text{DY}}=2}$                    ...(i)

Similarly, ΔAEQ ∼ ΔADY,
⇒ ${\displaystyle \frac{\text{EQ}}{\text{DY}}=\frac{\text{AE}}{\text{ED}}=\frac{1}{2}}$ ...( E is the mid-point of AD )

that is ${\displaystyle \frac{\text{EQ}}{\text{DY}}=\frac{1}{2}}$             ....(ii)

Dividing (i) by (ii), We get

⇒ ${\displaystyle \frac{\text{BQ}}{\text{EQ}}=4}$
⇒ BE + EQ = 4EQ
⇒ BE = 3EQ
⇒ ${\displaystyle \frac{\text{BQ}}{\text{EQ}}=\frac{3}{1}}$

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.

Sol:

ln ΔEDF,
M is the mid-point of AB and N is the mid-point of DE.
⇒ MN = ${\displaystyle \frac{1}{2}}$EF            ...( Mid-point theorem )
⇒ EF = 2MN                 ...(i)

ln ΔABC,
M is the mid-point  of AB and N is the mid-point of BC,
⇒ MN = ${\displaystyle \frac{1}{2}}$AC              ....( Mid-point theorem )

⇒ AC =2MN                     ....(ii)
From (i) and (ii), we get
⇒ EF = AC