# RS Aggarwal Solutions Class 9 Ex 9B

Q.1: In the adjoining figure ABCD is a trapezium in which ABDC$AB\parallel DC$ and E is the midpoint of AD. A line segment EFAB$EF \parallel AB$ meets BC at F. Show that F is the midpoint of BC.

Sol:

Join BD to cut EF at M

Now, in Δ$\Delta$DAB, E is the midpoint of AD and EM || AB

M is the midpoint of BD. (By converse of midpoint theorem)

Again, in Δ$\Delta$BDC, M is the midpoint of BD and MF || DC

Therefore, F is the midpoint of BC (By converse of midpoint theorem)

Q.2: In the adjoining figure ABCD is a parallelogram in which E and F are the midpoints of AB and CD respectively. If GH is a line segment that cuts AD, EF and BC at G, P and H respectively, prove that GP = PH.

Sol:

In parallelogram ABCD, we have:

AD || BC and AB || DC

AD = BC and AB = DC

AB = AE + BE and DC = DF + FC

AE = BE = DF = FC

Now, DF = AE and OF || AE.

i.e., AEFD is a parallelogram.

Similarly, BEFC is also a parallelogram.

EF || BC and AD || EF || BC

Thus, AD, EF and BC are three parallel lines cut by the transversal line DC at D, F and C, respectively such that DF = FC.

These lines AD, EF and BC are also cut by the transversal AB at A, E and B, respectively such that AE = BE

Similarly, they are also cut by GH

Therefore, GP = PH (By intercept theorem)

Q.3: In the adjoining figure ABCD is a trapezium in which ABDC$AB\parallel DC$ and P and Q are the midpoints of AD and BC respectively.  DQ and AB when produced meet at E. Also AC and PQ intersect at R . Prove that

(i) DQ = QE

(ii) PRAB$PR\parallel AB$

(iii) ARRC$AR\parallel RC$

Sol:

Given: AB || DC, AP = PD and BQ = CQ

(i) In Δ$\Delta$QCD and Δ$\Delta$QBE, we have:

∠DQC = ∠BQE (Vertically opposite angles)

∠DCQ = ∠EBQ (Alternate angles, as AE || DC)

BQ = CQ (P is the midpoints)

Δ$\Delta$QCD =Δ$\Delta$QBE

Hence, DQ = QE (CPCT)

(ii) Now, in Δ$\Delta$ADE, P and Q are the midpoints of AD and DE, respectively.

PQ || AE, PQ AB || DC and AB || PR || DC

(iii) PQ AB and DC are the three lines cut by transversal AD at P such that AP = PD. These lines PQ, AB, DC are also cut by transversal BC at Q such that BQ = QC. Similarly, lines PQ, AB and DC are also cut by AC at R. Therefore, AR = RC (By intercept theorem).

Q.4: In the adjoining figure AD is a median of ABC$\bigtriangleup ABC$ and DEBA$DE\parallel BA$.  Show that BE is also a median of triangle ABC$\bigtriangleup ABC$

Sol:

AD is a median of Δ$\Delta$ABC and BD = DC

We know that the line drawn through the midpoint of one side of a triangle and parallel to another side bisects the third side.

Here, in Δ$\Delta$ABC, D is the midpoint of BC and DE || BA (given). Then DE bisects AC.

i.e., AE = EC

E is the midpoint of AC

Therefore, BE is the median of Δ$\Delta$ ABC

Q.5: In the adjoining figure AD and BE are the medians of ABC$\bigtriangleup ABC$ and DFBE$DF\parallel BE$. Show that CF=14AC$CF=\frac{1}{4}AC$.

Sol:

In Δ$\Delta$ABC, we have:

AC = AE + EC . . . . . . (i)

AE = EC . . . . . . (ii) [BE is the median of ABDC]

AC = 2EC . . . . . . . (iii)

In Δ$\Delta$ BEC, DF || BE.

EF = CF (By midpoint theorem, as D is the midpoint of BC)

But EC = EF + CF

EC = 2 (CF) . . . . . . (iv)

From (iii) and (iv), we get:

AC = 2 (2CF)

Therefore, CF = 14 AC

Q.6: In the adjoining figure ABCD is a parallelogram, E is the midpoint of DC and through D, a line segment is drawn parallel to EB to meet CB produced at G and it cuts AB at F. Prove that:

(i). qAB=12GC$AB=\frac{1}{2}GC$

(ii). DG = 2 EB

Sol:

(i) In Δ$\Delta$ DCG, we have:

DG || EB and DE = EC (E is the midpoint of DC)

Also, GB = BC (By midpoint theorem)

B is the midpoint of GC

Now, GC = GB + BC

GC = 2BC = 2 (AD)

In ADCG, DG || EB and E is the midpoint of DC and B is the midpoint of GC.

By midpoint theorem, the line segment joining the mid points of any two sides of a triangle is parallel to the third side and is half of it.

i.e., EB = 12DG

DG = 2 (EB)

Q.7: Prove that the line segment joining the midpoints of the sides of a triangle divided it into 4 congruent triangles.

Sol:

Δ$\Delta$ ABC is shown in the figure

D, E and F are the midpoints of sides AB, BC and CA, respectively

As, D and E are the mid points of sides AB, and BC of Δ$\Delta$ ABC

DE || AC (By midpoint theorem)

Similarly, DF || BC and EF || AB

Therefore, ADEF, BDFE, and DFCE are all parallelograms.

Now, DE is the diagonal of the parallelogram BDFE.

Δ$\Delta$ BDE = Δ$\Delta$ FED

Similarly, DF is the diagonal of the parallelogram ADEF.

Δ$\Delta$ DAF = Δ$\Delta$ FED

And, EF is the diagonal of the parallelogram DFCE.

Δ$\Delta$ EFC = Δ$\Delta$ FED.

So, all the four triangles are congruent.

Q.8: In the adjoining figure D,E and F are the midpoints of the sides BC CA and AB respectively of ABC$\bigtriangleup ABC$. Show that EDF=A,DEF=BandDFE=C$\angle EDF = \angle A,\; \angle DEF= \angle B\; and\; \angle DFE= \angle C$.

Sol:

Δ$\Delta$ ABC is shown below. D, E and F are the midpoints of sides BC, CA and AB, respectively.

As F and E are the mid points of sides AB and AC of Δ$\Delta$

ABC.

FE || BC (By mid point theorem)

Similarly, DE || FB and FD || AC

Therefore, AFDE, BDEF, and DCEF are all parallelograms.

In parallelogram AFDE, we have:

∠A = ∠EDF (Opposite angles are equal)

In parallelogram BDEF, we have:

∠B = ∠DEF (Opposite angles are equal)

In parallelogram DCEF, we have:

∠C = ∠DFE (Opposite angles are equal)

Q.9: Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rectangle is a rhombus.

Sol:

Let, ABCD be the rectangle and P, Q, R and S be the midpoints of AB, BC, CD, and DA, respectively.

Join AC, a diagonal of the rectangle.
In Δ$\Delta$ ABC, we have:

PQ || AC and PQ = 12AC [By midpoint theorem]

Again, in Δ$\Delta$ DAC, the points S and R are the mid points of AD and DC, respectively.

SR || AC and SR = 12AC [By midpoint theorem]

Now, PQ || AC and SR || AC

PQ || SR

Also, PQ = SR [Each equal to 12 AC] . . . . . . . (i)

So, PQRS is a parallelogram.

Now, in Δ$\Delta$SAP and Δ$\Delta$QBP, we have:

AS = BQ, ∠A = ∠B = 90° and AP = BP

i.e., Δ$\Delta$ SAP = Δ$\Delta$ QBP

PS = PQ . . . . . . . . . (ii)

Similarly, ASDR = ACICR

SR = RQ . . . . . . . . (iii)

From (i), (ii) and (iii), we have:

PQ = PQ = SR = RQ

Hence, PQRS is a rhombus.

Q.10: Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a rhombus is a rectangle.

Sol:

Let, ABCD be a rhombus and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively. Join the diagonals, AC and BD.

In Δ$\Delta$ ABC, we have:

PQ || AC and PQ = 12AC [By midpoint theorem]

Again, in ADAC, the points S and R are the midpoints of AD and DC, respectively.

SR || AC and SR = 12AC [By midpoint theorem]

Now, PQ || AC and SR || AC   => PQ || SR

Also, PQ = SR [Each equal to 12 AC] . . . . . . . . . (i)

So, PQRS is a parallelogram.

We know that the diagonals of a rhombus bisect each other at right angles ∠EOF = 90°.

Now, RQ || DB => RE || FO

Also, SR || AC => FR || OE

Therefore, OERF is a parallelogram.

So, ∠FRE = ∠EOF = 90 (Opposite angles are equal)

Thus, PQRS is a parallelogram with ∠R = 90°.

Therefore, PQRS is a rectangle.

Q.11: Show that the quadrilateral formed by joining the midpoints of the pairs of adjacent sides of a square is a square

Sol:

Let, ABCD be a square and P, Q, R and S are the midpoints of AB, BC, CD and DA, respectively.

Join the diagonals AC and BD. Let BD cut SR at F and AC cut RQ at E. Let 0 be the intersection point of AC and BD.

In Δ$\Delta$ ABC, we have:

PQ || AC and PQ = 12AC [By midpoint theorem]

Again, in ADAC, the points S and R are the midpoints of AD and DC, respectively. SR || AC and SR = 12AC [By midpoint theorem]

Now, PQ || AC and SR || AC => PQ || SR

Also, PQ = SR     [Each equal to 12 AC]

So, PQRS is a parallelogram.

Now, in Δ$\Delta$ SAP and Δ$\Delta$ QBP, we have:

AS = BQ

AP = BP

i.e., Δ$\Delta$ SAP = Δ$\Delta$QBP

PS = PQ . . . . . . . . (ii)

Similarly, Δ$\Delta$SDR = Δ$\Delta$RCQ

SR = RQ . . . . . . . . . . (iii)

From (i), (ii) and (iii), we have:

PQ = PS = SR = RQ . . . . . . . . (iv)

We know that the diagonals of a square bisect each other at right angles.

∠EOF = 90°

Now, RQ || DB

=> RE || FO

Also, SR || AC

=>FR || OE

Therefore, OERF is a parallelogram.

So, ∠FRE = ∠EOF = 90° (Opposite angles are equal)

Thus, PQRS is a parallelogram with ∠R = 90° and PQ = PS = SR = RQ.

Therefore, PQRS is a square.

Q.12: Prove that the line segment joining the midpoints of opposite sides of a quadrilateral bisect each other.

Sol:

Let ABCD be the quadrilateral in which P, Q, R, and S are the midpoints of the sides AB, BC, CD and DA respectively.

Join PQ, QR, RS, SP and BD. BD is a diagonal of ABCD.

In Δ$\Delta$ABD, S and P are the midpoints of AD and AB, respectively. SP || BD and SP = 12$\frac{1}{2}$ BD … (i) (By midpoint theorem)

Similarly in Δ$\Delta$BCD. we have:

QR || BD and QR =12$\frac{1}{2}$  BD … (ii) (By midpoint theorem)

From equations (i) and (ii), we get:

SP || BD || QR

SP || QR and SP = OR   [Each equal to 12$\frac{1}{2}$  BD]

In quadrilateral SPQR, one pair of the opposite sides is equal and parallel to each other.

Therefore, SPQR is a parallelogram.

We know that the diagonals of a parallelogram bisect each other.

Therefore, PR and OS bisect each other.

Q.13: In the given figure, ABCD is a quadrilateral whose diagonals intersect at right angles. Show that the quadrilateral formed by joining the midpoints of the pair of adjacent it is a rectangle.

Sol:

ABCD is a quadrilateral and P, Q, R and S be the midpoints of AB, BC, CD and DA, respectively.

AC and BD are the diagonals which intersect each other at 0. RQ intersects AC at E and SR intersects BD at F.

In Δ$\Delta$ABC, we have:

PC || AC and PO = 12AC [By midpoint theorem]

Again, in ADAC, the points S and R are the midpoints of AD and DC, respectively.  SR || AC and SR = 12AC [By midpoint theorem]

Now, PC || AC and SR || AC

PQ || SR

Also, PQ = SR [Each equal to 12AC]

So, PARS is a parallelogram.

We know that the diagonals of the given quadrilateral bisect each other at right angles.

∠EOF = 90°

Now, RQ || DB

i.e. RE || FO

Also, SR || AC

i.e. FR || OE

OERF is a parallelogram.

So, ∠FRE = ∠EOF = 90° (Opposite angles are equal)

Thus, PQRS is a parallelogram with ∠R = 90°.

PQRS is a rectangle.

#### Practise This Question

Which of the following statements is not true about the figure given below?