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 ${\displaystyle \sqrt{3}:1}$