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

Mathematics in ICSE Class 9 is one of the most challenging and trickiest subjects of all. It includes complex topics such as logarithms, expansions, indices and Pythagoras Theorem which are difficult to understand for an average student. TopperLearning provides study materials for ICSE Class 9 Mathematics to make the subject easy and help students to clear all their concepts. Our study materials comprise numerous video lessons, question banks, revision notes and sample papers which help achieve success in the examination.

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

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

And

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 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

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.

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

The following figure shows atrapezium 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

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,

(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

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

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

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

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.

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

A parallelogram ABCD has P the mid-point of Dc and Q a point of Ac such that CQ = 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

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

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

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

Therefore QR=RS

Now

Hence proved

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)

(ii)For tiangle ECB PQ||CE

Again CE||AB

Therefore PQ||AB

Hence proved

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

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 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

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

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.

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

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=2BD==8.2

(iv)

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

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

The side AC of a triangle ABC is produced to point E so that CE = AC. 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

(i)

Again from the triangle *AEF* we have *AE *||*PD*||*CR* and

Therefore 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

Now

Hence it is shown.

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

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

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.

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)

(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.

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

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

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 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

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 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)=

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.

_{}

Substitute value of BE in first equation, we get

_{}

Hence Proved

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:

(i) BP = 2AD

(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.

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

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

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.

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.

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

OB^{2} + OC^{2} = BC^{2}

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