SELINA Solutions for Class 9 Maths Chapter 12 - Mid-point and Its Converse [ Including Intercept Theorem]

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Chapter 12 - Mid-point and Its Converse [ Including Intercept Theorem] Exercise Ex. 12(A)

Solution 1

The triangle is shown below,

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

And

Solution 2

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 and

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

Solution 3

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

Solution 4

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

Solution 5

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

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

From triangle ABC

From triangle ACD

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

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

From triangle ACD

Therefore ON=27-10=17cm

From triangle ABC

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

From triangle ABC

Therefore OM=15-11.5=3.5cm

From triangle ACD

Solution 6

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 intersects at right angle at point O. We need to show that PQRS is a rectangle

Proof:

From and

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

Again BD||RQ, therefore

Similarly

Therefore PQRS is a rectangle.

Hence proved

Solution 7

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

Solution 8

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

Solution 9

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,

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

Solution 10

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

Solution 11

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

Therefore QR=RS

Now

Hence proved

Solution 12

The required figure is shown below

(i)

(ii)For tiangle ECB PQ||CE

Again CE||AB

Therefore PQ||AB

Hence proved

Solution 13

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

Solution 14

The required figure is shown below

(i)Since F is the midpoint and EF||AB.

Therefore E is the midpoint of BC

So and …..(1)

Since D and F are the midpoint of AB and AC

Therefore DE||BC

SO and …..(2)

From (1),(2) we get

BE=DF and BD=EF

Hence BDEF is a parallelogram.

(ii)Since

Solution 15

Solution 16

Solution 17

  

Chapter 12 - Mid-point and Its Converse [ Including Intercept Theorem] Exercise Ex. 12(B)

Solution 1

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=2BD==8.2

(iv)

Solution 2

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

DS=DQ+QR+RS=QR+QR+RS=3RS

Solution 3

Consider the figure:

Here D is the midpoint of BC and DP is parallel to AB, therefore P is the midpoint of AC and Syntax error from line 1 column 49 to line 1 column 73. Unexpected 'mathsize'.

(i)

Again from the triangle AEF we have AE ||PD||CR and Syntax error from line 1 column 49 to line 1 column 73. Unexpected 'mathsize'.

Therefore Syntax error from line 1 column 49 to line 1 column 73. Unexpected 'mathsize'. 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 Syntax error from line 1 column 49 to line 1 column 73. Unexpected 'mathsize'.

Now

Syntax error from line 1 column 49 to line 1 column 73. Unexpected 'mathsize'.

Hence it is shown. 

Solution 4

The figure is shown below

(i)

From triangle BPC and triangle APN

And BC=AN ……(3)

Similarly …..(2)

And BC=AM ……(4)

Now

Therefore M,A,N are collinear

(ii) From (3) and (4) MA=NA

Hence A is the midpoint of MN

Solution 5

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.

Solution 6

The figure is shown below

(i)

(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 …..(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.

Solution 7

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

Solution 8

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

Solution 9

The figure is shown below

Let ABCD be a quadrilateral where P,Q,R,S are 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 ….(1)

||Again BD||RQ, therefore …(2)

From (1) and (2) we get ,

Similarly

Therefore diagonals AC and BD intersect at right angle

Hence proved

Solution 10

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 2DE=BC or DE=BF

Again D and E are midpoint ,therefore DE||BF and EF=BD

Hence BDEF is a parallelogram.

Now

Therefore perimeter of BDEF=2(BF+EF)=

Solution 11

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.

  

 

Substitute value of BE in first equation, we get

   

 

Hence Proved

Solution 12

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.


Hence BP = 2AD


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


Hence Proved

Solution 13

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

 

Solution 14

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.

 

Solution 15

If adjacent sides of a parallelogram are equal, then it is rhombus.

Now, the diagonals of a rhombus bisect each other and are perpendicular to each other.

Let the lengths of the diagonals be x and y.

Diagonal of length y be equal to the sides of rhombus.

Thus, each side of rhombus = y

  

Now, in right-angles ΔBOC, by Pythagoras theorem

OB2 + OC2 = BC2