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

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.

Sol:

According to equal intercept theorem since CD = DE
Therefore AB = BC and EF = GF
(i) BC = AB = 7.2cm
(ii) GE = EF + GF = 2EF = 2 x 4 =8cm

Since B,D,F are the mid-point and AE || BF || CG
Therefore AE = 2BD and CG = 2DF

(iii) AE = 2BD = 2 x 4 = 8.2
(iv) DF = ${\displaystyle \frac{1}{2}}$CG = ${\displaystyle \frac{1}{2}}$ x 11 = 5.5 cm

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.

Sol:

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

(i) From triangle DPR, A and Q are the mid-point 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 mid-point of BC

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

Question 3

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.

Sol:

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 = ${\displaystyle \frac{1}{2}}$AB

(i) Again from the triangle AEF, we have AE || PD || CR and AP = ${\displaystyle \frac{1}{3}}$AE

Therefore DF = ${\displaystyle \frac{1}{3}}$ 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 = ${\displaystyle \frac{1}{2}}$PD

Now,
PD = ${\displaystyle \frac{1}{2}}$ AB

${\displaystyle \frac{1}{2}\text{PD}=\frac{1}{4}}$AB

CR = ${\displaystyle \frac{1}{4}}$AB

4CR = AB
Hence it is shown.

Question 4

In triangle ABC, the medians BP and CQ are produced up to 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.

Sol:

The figure is shown below

(i) From triangle BPC and triangle APN
∠BPC = ∠APN                       ...[ Opposite angle ]
BP = AP
PC = PN
∴ ΔBPC ≅ ΔAPN                  ...[ 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.

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.

Sol:

The figure is shown below

From figure EF || AB and E is the mid-point 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.

Sol:

The figure is shown below

(i) From ΔHEB and ΔFHC
BE = FC
∠EHB = ∠FHC                        ...[ Opposite angle ]
∠HBE = ∠HFC
∴  ΔHEB ≅ ΔFHC
∴  EH = CH , BH = FH

(ii) Similarly AG = GF and EG = DG      …..(1)
For triangle ECD,
F and H are the mid-point of CD and EC.
Therefore HF || DE and 
HF = ${\displaystyle \frac{1}{2}}$ DE                                          ....(2)

From (1) and (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 meets side BC at points M and N respectively. Prove that: BM = MN = NC.

SOl:

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 the intercept Theorem to show that MN bisects AD.

Sol:

The figure is shown below

Since M and N are the mid-point 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 the right angle.

Sol:

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 a right angle.

Proof:
PQ || AC, therefore ∠AOD = ∠PXO     ...[ Corresponding angle ]...(1)

Again BD || RQ, therefore ∠PXO = ∠RQX = 90°  ....[ Corresponding angle and angle of a 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.

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.

SoL:

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 midpoints,
therefore DE || BF and EF = BD
Hence BDEF is a parallelogram.
Now,
BD = EF = ${\displaystyle \frac{1}{2}\text{AB}=\frac{1}{2}}$ x 16 = 8 cm

BF = DE = ${\displaystyle \frac{1}{2}\text{BC}=\frac{1}{2}}$ x 18 = 9 cm

Therefore perimeter of BDEF = 2( BF + EF ) = 2( 9 + 8 ) = 34 cm.

Question 11

In the given figure, AD and CE are medians and DF // CE.
Prove that: FB = ${\displaystyle \frac{1}{4}}$ AB.

Sol:

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 the 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 the midpoint of BC.
So F must be the midpoint of BE.
So, FB = ${\displaystyle \frac{1}{2}}$BE but BE = ${\displaystyle \frac{1}{2}}$AB

Substitute value of BE in the first equation, we get
FB = ${\displaystyle \frac{1}{4}}$AB
Hence Proved.

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:
(i) BP = 2AD
(ii) O is the mid-point of AP.

Sol:

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

Consider triangle APB, given EC, is parallel to AP and E is the 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.

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.

Sol:

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

We know that AB = GH = IJ
From midpoint theorem,

EG = ${\displaystyle \frac{1}{2}\text{DI}, \text{HF}=\frac{1}{2}}$JC

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.

Sol:

Since AD is the median of ΔABC, then BD = DC.

Given, DE || AB and DE are drawn from the midpoint of BC i.e. D, then

by the converse of mid-point theorem,

it bisects the third side which in this case is AC at E.

Therefore, E is the mid point of AC.

Hence, BE is the median of ΔABC.

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 ratio √3:1.

Sol:

If adjacent sides of a parallelogram are equal, then it is a 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 the rhombus.
Thus, each side of rhombus = y

Now, in right-angles ΔBOC, by Pythagoras theorem
OB² + OC² = BC²
⇒ ${\displaystyle {\left(\frac{y}{2}\right)}^2+{\left(\frac{x}{2}\right)}^2=y^2}$

⇒ ${\displaystyle \left(\frac{x^2}{4}\right)=y^2-\frac{y^2}{4}}$

⇒ ${\displaystyle \left(\frac{x^2}{4}\right)=\frac{4y^2-y^2}{4}}$

⇒ ${\displaystyle \left(\frac{x^2}{4}\right)=\frac{3y^2}{4}}$

⇒ ${\displaystyle x^2=3y^2}$

⇒ ${\displaystyle \frac{x^2}{y^2}=\frac{3}{1}}$

⇒ ${\displaystyle \frac{x}{y}=\frac{\sqrt{3}}{1}}$

Thus, the diagonal are in the ratio ${\displaystyle \sqrt{3}:1}$

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