### Exercise 8.1

**Question 1. The angles of quadrilateral are in the ratio 3 : 5 : 9 : 13. Find all the angles of the quadrilateral.**

**Solution:**

Let the angles of the quadrilateral be 3x, 5x, 9x and 13x.

∴ 3x + 5x + 9x + 13x = 360°

[Angle sum property of a quadrilateral]

⇒ 30x = 360°

⇒ x = = 12°

∴ 3x = 3 x 12° = 36°

5x = 5 x 12° = 60°

9x = 9 x 12° = 108°

13a = 13 x 12° = 156°

⇒ The required angles of the quadrilateral are 36°, 60°, 108° and 156°.

**Question 2. If the diagonals of a parallelogram are equal, then show that it is a rectangle.****Solution:**

Let ABCD is a parallelogram such that AC = BD.

In ∆ABC and ∆DCB,

AC = DB [Given]

AB = DC [Opposite sides of a parallelogram]

BC = CB [Common]

∴ ∆ABC ≅ ∆DCB [By SSS congruency]

⇒ ∠ABC = ∠DCB [By C.P.C.T.] …(1)

Now, AB || DC and BC is a transversal. [ ∵ ABCD is a parallelogram]

∴ ∠ABC + ∠DCB = 180° … (2) [Co-interior angles]

From (1) and (2), we have

∠ABC = ∠DCB = 90°

i.e., ABCD is a parallelogram having an angle equal to 90°.

∴ ABCD is a rectangle.

**Question 3. Show that if the diagonals of a quadrilateral bisect each other at right angles, then it is a rhombus.****Solution:**

Let ABCD be a quadrilateral such that the diagonals AC and BD bisect each other at right angles at O.

∴ In ∆AOB and ∆AOD, we have

AO = AO [Common]

OB = OD [O is the mid-point of BD]

∠AOB = ∠AOD [Each 90]

∴ ∆AQB ≅ ∆AOD [By,SAS congruency

∴ AB = AD [By C.P.C.T.] ……..(1)

Similarly, AB = BC .. .(2)

BC = CD …..(3)

CD = DA ……(4)

∴ From (1), (2), (3) and (4), we have

AB = BC = CD = DA

Thus, the quadrilateral ABCD is a rhombus.

Alternatively : ABCD can be proved first a parallelogram then proving one pair of adjacent sides equal will result in rhombus.

**Question 4. Show that the diagonals of a square are equal and bisect each other at right angles.****Solution:**

Let ABCD be a square such that its diagonals AC and BD intersect at O.

(i) To prove that the diagonals are equal, we need to prove AC = BD.

In ∆ABC and ∆BAD, we have

AB = BA [Common]

BC = AD [Sides of a square ABCD]

∠ABC = ∠BAD [Each angle is 90°]

∴ ∆ABC ≅ ∆BAD [By SAS congruency]

AC = BD [By C.P.C.T.] …(1)

(ii) AD || BC and AC is a transversal. [∵ A square is a parallelogram]

∴ ∠1 = ∠3

[Alternate interior angles are equal]

Similarly, ∠2 = ∠4

Now, in ∆OAD and ∆OCB, we have

AD = CB [Sides of a square ABCD]

∠1 = ∠3 [Proved]

∠2 = ∠4 [Proved]

∴ ∆OAD ≅ ∆OCB [By ASA congruency]

⇒ OA = OC and OD = OB [By C.P.C.T.]

i.e., the diagonals AC and BD bisect each other at O. …….(2)

(iii) In ∆OBA and ∆ODA, we have

OB = OD [Proved]

BA = DA [Sides of a square ABCD]

OA = OA [Common]

∴ ∆OBA ≅ ∆ODA [By SSS congruency]

⇒ ∠AOB = ∠AOD [By C.P.C.T.] …(3)

∵ ∠AOB and ∠AOD form a linear pair.

∴∠AOB + ∠AOD = 180°

∴∠AOB = ∠AOD = 90° [By(3)]

⇒ AC ⊥ BD …(4)

From (1), (2) and (4), we get AC and BD are equal and bisect each other at right angles.

**Question 5. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.****Solution:**

Let ABCD be a quadrilateral such that diagonals AC and BD are equal and bisect each other at right angles.

Now, in ∆AOD and ∆AOB, We have

∠AOD = ∠AOB [Each 90°]

AO = AO [Common]

OD = OB [ ∵ O is the midpoint of BD]

∴ ∆AOD ≅ ∆AOB [By SAS congruency]

⇒ AD = AB [By C.P.C.T.] …(1)

Similarly, we have

AB = BC … (2)

BC = CD …(3)

CD = DA …(4)

From (1), (2), (3) and (4), we have

AB = BC = CD = DA

∴ Quadrilateral ABCD have all sides equal.

In ∆AOD and ∆COB, we have

AO = CO [Given]

OD = OB [Given]

∠AOD = ∠COB [Vertically opposite angles]

So, ∆AOD ≅ ∆COB [By SAS congruency]

∴∠1 = ∠2 [By C.P.C.T.]

But, they form a pair of alternate interior angles.

∴ AD || BC

Similarly, AB || DC

∴ ABCD is a parallelogram.

∴ Parallelogram having all its sides equal is a rhombus.

∴ ABCD is a rhombus.

Now, in ∆ABC and ∆BAD, we have

AC = BD [Given]

BC = AD [Proved]

AB = BA [Common]

∴ ∆ABC ≅ ∆BAD [By SSS congruency]

∴ ∠ABC = ∠BAD [By C.P.C.T.] ……(5)

Since, AD || BC and AB is a transversal.

∴∠ABC + ∠BAD = 180° .. .(6) [ Co – interior angles]

⇒ ∠ABC = ∠BAD = 90° [By(5) & (6)]

So, rhombus ABCD is having one angle equal to 90°.

Thus, ABCD is a square.

**Question 6. Diagonal AC of a parallelogram ABCD bisects ∠A (see figure). Show that(i) it bisects ∠C also,(ii) ABCD is a rhombus.**

**Solution:**

We have a parallelogram ABCD in which diagonal AC bisects ∠A

⇒ ∠DAC = ∠BAC

(i) Since, ABCD is a parallelogram.

∴ AB || DC and AC is a transversal.

∴ ∠1 = ∠3 …(1)

[ ∵ Alternate interior angles are equal]

Also, BC || AD and AC is a transversal.

∴ ∠2 = ∠4 …(2)

[ v Alternate interior angles are equal]

Also, ∠1 = ∠2 …(3)

[ ∵ AC bisects ∠A]

From (1), (2) and (3), we have

∠3 = ∠4

⇒ AC bisects ∠C.

(ii) In ∆ABC, we have

∠1 = ∠4 [From (2) and (3)]

⇒ BC = AB …(4)

[ ∵ Sides opposite to equal angles of a ∆ are equal]

Similarly, AD = DC ……..(5)

But, ABCD is a parallelogram. [Given]

∴ AB = DC …(6)

From (4), (5) and (6), we have

AB = BC = CD = DA

Thus, ABCD is a rhombus.

**Question 7. ABCD is a rhombus. Show that diagonal AC bisects ∠Aas well as ∠C and diagonal BD bisects ∠B as well AS ∠D.****Solution:**

Since, ABCD is a rhombus.

⇒ AB = BC = CD = DA

Also, AB || CD and AD || BC

Now, CD = AD ⇒ ∠1 = ∠2 …….(1)

[ ∵ Angles opposite to equal sides of a triangle are equal]

Also, AD || BC and AC is the transversal.

[ ∵ Every rhombus is a parallelogram]

⇒ ∠1 = ∠3 …(2)

[ ∵ Alternate interior angles are equal]

From (1) and (2), we have

∠2 = ∠3 …(3)

Since, AB || DC and AC is transversal.

∴ ∠2 = ∠4 …(4)

[ ∵ Alternate interior angles are equal] From (1) and (4),

we have ∠1 = ∠4

∴ AC bisects ∠C as well as ∠A.

Similarly, we can prove that BD bisects ∠B as well as ∠D.

**Question 8. ABCD is a rectangle in which diagonal AC bisects ∠A as well as ∠C. Show that(i) ABCD is a square(ii) diagonal BD bisects ∠B as well as ∠D.**

**Solution:**

We have a rectangle ABCD such that AC bisects ∠A as well as ∠C.

i.e., ∠1 = ∠4 and ∠2 = ∠3 ……..(1)

(i) Since, every rectangle is a parallelogram.

∴ ABCD is a parallelogram.

⇒ AB || CD and AC is a transversal.

∴∠2 = ∠4 …(2)

[ ∵ Alternate interior angles are equal]

From (1) and (2), we have

∠3 = ∠4

In ∆ABC, ∠3 = ∠4

⇒ AB = BC

[ ∵ Sides opposite to equal angles of a A are equal]

Similarly, CD = DA

So, ABCD is a rectangle having adjacent sides equal.

⇒ ABCD is a square.

(ii) Since, ABCD is a square and diagonals of a square bisect the opposite angles.

So, BD bisects ∠B as well as ∠D.

**Question 9. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see figure). Show that****Solution:**

We have a parallelogram ABCD, BD is the diagonal and points P and Q are such that PD = QB

(i) Since, AD || BC and BD is a transversal.

∴ ∠ADB = ∠CBD [ ∵ Alternate interior angles are equal]

⇒ ∠ADP = ∠CBQ

Now, in ∆APD and ∆CQB, we have

AD = CB [Opposite sides of a parallelogram ABCD are equal]

PD = QB [Given]

∠ADP = ∠CBQ [Proved]

∴ ∆APD ≅ ∆CQB [By SAS congruency]

(ii) Since, ∆APD ≅ ∆CQB [Proved]

⇒ AP = CQ [By C.P.C.T.]

(iii) Since, AB || CD and BD is a transversal.

∴ ∠ABD = ∠CDB

⇒ ∠ABQ = ∠CDP

Now, in ∆AQB and ∆CPD, we have

QB = PD [Given]

∠ABQ = ∠CDP [Proved]

AB = CD [ Y Opposite sides of a parallelogram ABCD are equal]

∴ ∆AQB = ∆CPD [By SAS congruency]

(iv) Since, ∆AQB = ∆CPD [Proved]

⇒ AQ = CP [By C.P.C.T.]

(v) In a quadrilateral ∆PCQ,

Opposite sides are equal. [Proved]

∴ ∆PCQ is a parallelogram.

**Question 10. ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see figure). Show that****Solution:**

(i) In ∆APB and ∆CQD, we have

∠APB = ∠CQD [Each 90°]

AB = CD [ ∵ Opposite sides of a parallelogram ABCD are equal]

∠ABP = ∠CDQ

[ ∵ Alternate angles are equal as AB || CD and BD is a transversal]

∴ ∆APB = ∆CQD [By AAS congruency]

(ii) Since, ∆APB ≅ ∆CQD [Proved]

⇒ AP = CQ [By C.P.C.T.]

**Question 11. In ∆ABC and ∆DEF, AB = DE, AB || DE, BC – EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F, respectively (see figure).Show that(i) quadrilateral ABED is a parallelogram(ii) quadrilateral BEFC is a parallelogram**

(iii) AD || CF and AD = CF

(iv) quadrilateral ACFD is a parallelogram

(v) AC = DF

(vi) ∆ABC ≅ ∆DEF

**Solution:**

(i) We have AB = DE [Given]

and AB || DE [Given]

i. e., ABED is a quadrilateral in which a pair of opposite sides (AB and DE) are parallel and of equal length.

∴ ABED is a parallelogram.

(ii) BC = EF [Given]

and BC || EF [Given]

i.e. BEFC is a quadrilateral in which a pair of opposite sides (BC and EF) are parallel and of equal length.

∴ BEFC is a parallelogram.

(iii) ABED is a parallelogram [Proved]

∴ AD || BE and AD = BE …(1)

[ ∵ Opposite sides of a parallelogram are equal and parallel] Also, BEFC is a parallelogram. [Proved]

BE || CF and BE = CF …(2)

[ ∵ Opposite sides of a parallelogram are equal and parallel]

From (1) and (2), we have

AD || CF and AD = CF

(iv) Since, AD || CF and AD = CF [Proved]

i.e., In quadrilateral ACFD, one pair of opposite sides (AD and CF) are parallel and of equal length.

∴Quadrilateral ACFD is a parallelogram.

(v) Since, ACFD is a parallelogram. [Proved]

So, AC =DF [∵ Opposite sides of a parallelogram are equal]

(vi) In ∆ABC and ∆DFF, we have

AB = DE [Given]

BC = EF [Given]

AC = DE [Proved in (v) part]

∆ABC ≅ ∆DFF [By SSS congruency]

**Question 12. ABCD is a trapezium in which AB || CD and AD = BC (see figure). Show that**

(i )∠A=∠B

(ii )∠C=∠D

(iii) ∆ABC ≅ ∆BAD

(iv) diagonal AC = diagonal BD

[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E].**Solution:**

We have given a trapezium ABCD in which AB || CD and AD = BC.

(i) Produce AB to E and draw CF || AD.. .(1)

∵ AB || DC

⇒ AE || DC Also AD || CF

∴ AECD is a parallelogram.

⇒ AD = CE …(1)

[ ∵ Opposite sides of the parallelogram are equal]

But AD = BC …(2) [Given]

By (1) and (2), BC = CF

Now, in ∆BCF, we have BC = CF

⇒ ∠CEB = ∠CBE …(3)

[∵ Angles opposite to equal sides of a triangle are equal]

Also, ∠ABC + ∠CBE = 180° … (4)

[Linear pair]

and ∠A + ∠CEB = 180° …(5)

[Co-interior angles of a parallelogram ADCE]

From (4) and (5), we get

∠ABC + ∠CBE = ∠A + ∠CEB

⇒ ∠ABC = ∠A [From (3)]

⇒ ∠B = ∠A …(6)

(ii) AB || CD and AD is a transversal.

∴ ∠A + ∠D = 180° …(7) [Co-interior angles]

Similarly, ∠B + ∠C = 180° … (8)

From (7) and (8), we get

∠A + ∠D = ∠B + ∠C

⇒ ∠C = ∠D [From (6)]

(iii) In ∆ABC and ∆BAD, we have

AB = BA [Common]

BC = AD [Given]

∠ABC = ∠BAD [Proved]

∴ ∆ABC = ∆BAD [By SAS congruency]

(iv) Since, ∆ABC = ∆BAD [Proved]

⇒ AC = BD [By C.P.C.T.]

### NCERT Solutions for Class 9 Maths Exercise 8.2

**Question 1. ABCD is a quadrilateral in which P, Q, R and S are the mid-points of sides AB, BC, CD and DA respectively (See figure). AC is a diagonal. Show that :**

(i) SR AC and SR = AC

(ii) PQ = SR

(iii) PQRS is a parallelogram.

**Solution:**

In ABC, P is the mid-point of AB and Q is the mid-point of BC.

Then PQ AC and PQ = AC

(i) In ACD, R is the mid-point of CD and S is the mid-point of AD.

Then SR AC and SR = AC

(ii) Since PQ = AC and SR = AC

Therefore, PQ = SR

(iii) Since PQ AC and SR AC

Therefore, PQ SR [two lines parallel to given line are parallel to each other]

Now PQ = SR and PQ SR

Therefore, PQRS is a parallelogram.

**Question 2. ABCD is a rhombus and P, Q, R, S are mid-points of AB, BC, CD and DA respectively. Prove that quadrilateral PQRS is a rectangle.**

**Solution:**

Given: P, Q, R and S are the mid-points of respective sides AB, BC, CD and DA of rhombus. PQ, QR, RS and SP are joined.

To prove: PQRS is a rectangle.

Construction: Join A and C.

Proof: In ABC, P is the mid-point of AB and Q is the mid-point of BC.

PQ AC and PQ = AC ……….(i)

In ADC, R is the mid-point of CD and S is the mid-point of AD.

SR AC and SR = AC ……….(ii)

From eq. (i) and (ii), PQ SR and PQ = SR

PQRS is a parallelogram.

Now ABCD is a rhombus. [Given]

AB = BC

AB = BC PB = BQ

1 = 2 [Angles opposite to equal sides are equal]

Now in triangles APS and CQR, we have,

AP = CQ [P and Q are the mid-points of AB and BC and AB = BC]

Similarly, AS = CR and PS = QR [Opposite sides of a parallelogram]

APS CQR [By SSS congreuancy]

3 = 4 [By C.P.C.T.]

Now we have 1 + SPQ + 3 =

And 2 + PQR + 4 = [Linear pairs]

1 + SPQ + 3 = 2 + PQR + 4

Since 1 = 2 and 3 = 4 [Proved above]

SPQ = PQR ……….(iii)

Now PQRS is a parallelogram [Proved above]

SPQ + PQR = ……….(iv) [Interior angles]

Using eq. (iii) and (iv),

SPQ + SPQ = 2SPQ =

SPQ =

Hence PQRS is a rectangle.

**Question 3. ABCD is a rectangle and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.**

**Solution:Given: A rectangle ABCD in which P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. PQ, QR, RS and SP are joined.**

**To prove: PQRS is a rhombus.**

**Construction: Join AC.**

**Proof: In ABC, P and Q are the mid-points of sides AB, BC respectively.**

**PQ AC and PQ = AC ……….(i)**

**In ADC, R and S are the mid-points of sides CD, AD respectively.**

**SR AC and SR = AC ……….(ii)**

**From eq. (i) and (ii), PQ SR and PQ = SR ……….(iii)**

**PQRS is a parallelogram.**

**Now ABCD is a rectangle. [Given]**

**AD = BC**

**AD = BC AS = BQ ……….(iv)**

**In triangles APS and BPQ,**

**AP = BP [P is the mid-point of AB]**

**PAS = PBQ [Each ]**

**And AS = BQ [From eq. (iv)]**

**APS BPQ [By SAS congruency]**

**PS = PQ [By C.P.C.T.] ………(v)**

**From eq. (iii) and (v), we get that PQRS is a parallelogram.**

**PS = PQ**

**Two adjacent sides are equal.**

**Hence, PQRS is a rhombus.**

**Question 4. ABCD is a trapezium, in which AB DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E, parallel to AB intersecting BC at F (See figure). Show that F is the mid-point of BC.**

**Solution:Let diagonal BD intersect line EF at point P.**

**In DAB,**

**E is the mid-point of AD and EP AB [ EF AB (given) P is the part of EF]**

**P is the mid-point of other side, BD of DAB.**

**[A line drawn through the mid-point of one side of a triangle, parallel to another side intersects the third side at the mid-point]**

**Now in BCD,**

**P is the mid-point of BD and PF DC [EF AB (given) and AB DC (given)]**

**EF DC and PF is a part of EF.**

**F is the mid-point of other side, BC of BCD. [Converse of mid-point of theorem]**

**Question 5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (See figure). Show that the line segments AF and EC trisect the diagonal BD.**

**Solution:Since E and F are the mid-points of AB and CD respectively.**

**AE = AB and CF = CD……….(i)**

**But ABCD is a parallelogram.**

**AB = CD and AB DC**

**AB = CD and AB DC**

**AE = FC and AE FC [From eq. (i)]**

**AECF is a parallelogram.**

**FA CE FP CQ [FP is a part of FA and CQ is a part of CE] ………(ii)**

**Since the segment drawn through the mid-point of one side of a triangle and parallel to the other side bisects the third side.**

**In DCQ, F is the mid-point of CD and FP CQ**

**P is the mid-point of DQ.**

**DP = PQ ……….(iii)**

**Similarly, In ABP, E is the mid-point of AB and EQ AP**

**Q is the mid-point of BP.**

**BQ = PQ ……….(iv)**

**From eq. (iii) and (iv),**

**DP = PQ = BQ ………(v)**

**Now BD = BQ + PQ + DP = BQ + BQ + BQ = 3BQ**

**BQ = BD ……….(vi)**

**From eq. (v) and (vi),**

**DP = PQ = BQ = BD**

**Points P and Q trisects BD.**

**So AF and CE trisects BD.**

**Question 6. Show that the line segments joining the mid-points of opposite sides of a quadrilateral bisect each other.**

**Solution:Given: A quadrilateral ABCD in which EG and FH are the line-segments joining the mid-points of opposite sides of a quadrilateral.**

**To prove: EG and FH bisect each other.**

**Construction: Join AC, EF, FG, GH and HE.**

**Proof: In ABC, E and F are the mid-points of respective sides AB and BC.**

**EF AC and EF AC ……….(i)**

**Similarly, in ADC,**

**G and H are the mid-points of respective sides CD and AD.**

**HG AC and HG AC ……….(ii)**

**From eq. (i) and (ii),**

**EF HG and EF = HG**

**EFGH is a parallelogram.**

**Since the diagonals of a parallelogram bisect each other, therefore line segments (i.e. diagonals) EG and FH (of parallelogram EFGH) bisect each other.**

**Question 7. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D.**

**Solution: (i) In ABC, M is the mid-point of AB [Given]**

**MD BC**

**AD = DC [Converse of mid-point theorem]**

**Thus D is the mid-point of AC.**

**(ii) BC (given) consider AC as a transversal.**

**1 = C [Corresponding angles]**

**1 = [C = ]**

**Thus MD AC.**

**(iii) In AMD and CMD,**

**AD = DC [proved above]**

**1 = 2 = [proved above]**

**MD = MD [common]**

**AMD CMD [By SAS congruency]**

**AM = CM [By C.P.C.T.] ……….(i)**

**Given that M is the mid-point of AB.**

**AM = AB ……….(ii)**

**From eq. (i) and (ii),**

**CM = AM = AB**