Class 9 Maths Chapter 8 Quadrilaterals

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.
Given, the ratio of the angles of quadrilateral are 3:5:9:13.
Let the angles of the quadrilateral are 3x, 5x, 9x and 13x.
We know that, sum of angles of a quadrilateral = 360°
∴ 3x + 5x + 9x +13x = 360°
∴ 30x = 360° ⇒ °
∴ Angles of the quadrilateral are
3x = 3 x 12 = 36°
5x = 5 x 12 = 60°
9x = 9 x 12 = 108°
and 13x =13 x 12=156°

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

We have a parallelogram ABCD such that AC = BD.
In Δ ABC and Δ DCB
AC = DB [given]
AB = DC (opposite sides of a parallogram)
BC=CB (Common)
Δ ABC ≅ Δ DCB  [by SAS congrunce rule]
∠ABC ≅ ∠DCB  ………(i)
∵ AB || DC and BC is a transversal. [ ∵ ABCD is a parallelogram]
∴ ∠ABC + ∠DCB =180°
(interior  opposite angles are supplementry]
From eqs. (i) and (ii), we have
∠ABC = ∠DCB =90°
i.e., ABCD is a parallelogram having an angle equal to 90°.
∴ ABCD is a rectangle.
Hence proved.

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

We have a quadrilateral ABCD such that the diagonals AC and BD bisect each other at right angles at O.
∴  In Δ AOB, and Δ AOD, we have
AO = AO common]
OB = OD [∵ O is the mid-point of BD]
∠AOB = ∠AOD [each = 90 °]
Δ AOB ≅  Δ AOD [by SAS congruence rule]
AB = AD [by CPCT]… (i)
Similarly, AB = BC …(ii)
BC = CD …(iii)
CD = AD …(iv)
∴  From eqs.(i), (ii), (iii) and (iv), we have AB = BC = CD = DA Hence, the quadrilateral ABCD is a rhombus.

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

Given: A square PQRS in which diagonals PR and QS intersect each other at 0.
To prove : PR = SQ, PO = OR, QO = OS and PR ⊥ SQ.
Proof: In Δ POQ and Δ ROS
PQ = RS [sides of a square]
∠PQO=∠RSO [alternative interior angles]
∠QPO = ∠SRO [alternative interior angles]
⇒ Δ POQ ≅ Δ ROS [by SAS congruence rule]
⇒   PO = OR [by CPCT]
and QO = OS [by CPCT]
Thus, PR and SQ bisect each other at O.
Also, in Δ POQ and Δ ROQ,
PQ = RQ
PO = OR
OQ = OQ
⇒ ΔPOQ ≅ ΔROQ by SAS congruence rule]
⇒ ∠POQ = ∠ROQ
Also.,∠POQ + ∠ROQ = 180°
⇒ ∠POQ + ∠POQ = 180°
⇒2 ∠POQ = 180°
⇒ ∠POQ = x 180° = 90°
Thus, diagonals PR and SQ are perpendicular to each other.
Again, in Δ PQS and Δ QPR
PQ = PQ
PS = QR
∠QPS – ∠PQR
⇒ ΔPQS = ΔQPR
⇒   SQ = PR
Hence, diagonals of a square 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.
Given: A quadrilateral ABCD in which AC = BD and AC⊥ BD such that OA = OC and OB = OD. So, ABCD is a parallelogram.

To prove: ABCD is a square.
Proof: Let AC and BD intersect at a point O.
In Δ ABO and Δ ADO, we have
BO = OD (Given)
AO = OA  (Common)
∴  ∠AOB = ∠AOD = 90° (Given)
∴  ΔABO ≅ ΔADO  (By SAS)
AB = AD  (By CPCT)
Also,AB = DC
and AD = BC (Opposite sides of parallelogram)
∴ AB = BC = DC = AD    .(i)
Again, in Δ ABC and Δ BAD, we have AB – BA AC = BD BC – AD
∴  Δ ABC = Δ BAD
∴  ∠ABC = ∠BAD…..(ii)
But ∠ABC + ∠BAD = 180°(Sum of interior angles of a parallelogram)
∴  ∠ABC = ∠BAD = 90° [From Eq. (ii)]
Thus, AB = BC = CD = DA and ∠A = 90°
∴  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.

(i) Given, diagonal AC of a parallelogram ABCD bisects ∠A.
i.e., ∠DAC =∠BAC =  ∠BAD           … (i)
Here, AB || DC and AC is a transversal.
∴ ∠DCA = ∠BAC               …(ii)[pair of alternate angles]
Similarly, BC || AD and BD is a transversal.
 ∴ ∠BCA = ∠DAC               … (iii) [pair of alternate angles]
From eqs. (i), (ii) and (iii), we get
∠DAC = ∠BCA = ∠BAC = ∠DCA .. (iv)
Now, ∠BCD = ∠BCA + ∠DCA
= ∠DAC + ∠BAC = ∠BAD
Hence, diagonal AC also bisects ∠C.

(ii) From eq. (iv), ∠DAC = ∠DCA
⇒ CD = DA [∵ angles opposite to equal sides are equal]
But AB = DC and AD =BC  [∵ ABCD is a parallelogram]
∴ AB = BC = CD = DA
Hence, ABCD is a rhombus.

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

Given : ABCD is a rhombus in which AB = BC =CD = AD.
To prove :
(i) Diagonal AC bisects ∠A as well as ∠C.

(ii) Diagonal BD bisects ∠B as well as
Construction: Join diagonals AC and BD.
Proof : ABCD is a rhombus.
∴  AB = BC = CD = AD
Also, AB || CD and AD || BC
Now,AD = CD ⇒ ∠1=∠2 …(i) [angles opposite to equal sides are equal]
Also, CD || AB  [opposite sides of the parallelogram]
and AC is transversal. ∠1 = ∠3    … (ii) [alternate interior angles]
From eqs. (i) and (ii), we have
∠2 = ∠3 and ∠1 = ∠4 ⇒ AC bisects ∠C as well as ∠A.
Similarly, we 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 ZB as well as
Solution.

We have a rectangle ABCD such that AC bisects ∠A as well as ∠C.
i.e. ∠1 = 4 and ∠2 = ∠3 …(i)
(i) Since rectangle is a parallelogram.
∴  ABCD is a parallelogram.
⇒ AS || CD and AC is a transversal.
 ∴ ∠2=∠4              … (ii) [alternate interior angles]
From eqs. (i) and (ii), we have
∠3 = ∠4 ⇒ AB = BC [∵ sides opposite to equal angles in Δ ABC are equal]
∴ AB = BC = CD = AD
⇒ ABCD is a rectangle having all of its sides equal.
∴ ABCD is a square.

(ii) Since ABCD is a square, and diagonals of a square bisect the opposite
 ∴ BD bisects ZB as well as ZD.

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

(i) Δ APD = Δ CQB
(ii) AP = CQ
(iii)  Δ AQB ≅ Δ  CPD
(iv) AQ = CP
(v) APCQis a parallelogram.
Solution.

Now, these equal angles form a pair of alternate angle when line segment AP and QC are intersected by a transversal PQ.
 ∴AP || CQ and AQ || CP
Now, both pairs of opposite sides of quadrilateral Δ PCQ are parallel. Hence, ΔPCQ is a parallelogram.  Hence proved.

Question 10.
ABCD is a parallelogram and AP and CQ are perpendiculars from vertices A and C on diagonal BD (see figure). Show that
(i) Δ APB ≅ Δ CQD
(ii) AP = CQ

Solution.
(i) In Δ APB and Δ CQD, we have
∠ APB = ∠CQD  [each = 90 °]
AB = CD  [opposite sides of parallelogram ABCD]
∠ABP= ∠CDQ
Δ APB = Δ CQD

(ii) Since, Δ APB = Δ CQD
∴  Their corresponding parts are equal.
AP = CQ

Question 11.
In ΔABC and ΔDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joined to vertices D, E and F respectively (see the 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) ΔABC ≅ ΔDEF

Solution.
Given, in Δ ABC and Δ DEF,
AB = DE,AB || DE, BC = EF and BC || EF
(i) Now, in quadrilateral ABED,
AB = DE and AB || DE  [given]
So, ABED is a parallelogram. [∵ a pair of opposite side is equal and parallel]

(ii) In quadrilateral BEFC,
BC = EF and BC || EF   [given]
So, BEFC is a parallelogram. [∵ a pair of opposite sides is equal and parallel]

(iii) Since, ABED is a parallelogram.
∴ AD || BE and AD = BE     …(i)
Also, BEFC is a parallelogram.
∴  CF || BE and CF = BE   …(ii)
From eqs. (i) and (ii), we get
AD || CF and AD = CF

(iv) In quadrilateral ACFD
∴  AD || CF and AD = CF [from part (iii)]
So, ACFD is a parallelogram.
Since, a pair of opposite sides is equal and parallel.

(v) Since, ACFD is a parallelogram.
∴ AC = DF and AC 11 DF

(vi) Now, in Δ ABC and Δ DEF,
AB = DE  [given]
BC = EF [given]
and  AC = DF [from part (v)]
∴ Δ ABC ≅ Δ DEF [by SSS congruence rule]
Hence proved.

Question 12.
ABCD is a trapezium in which AB || CD and AD = BC (see the figure). Show that
(i) ∠A = ∠B
(ii) ∠C = ∠D
(iii) Δ ABC = Δ BAD
(iv) Diagonal AC = Diagonal
[Hint : Extend AB and draw a line through C parallel to DA . intersecting AB produced at E.]

Solution.

Exercise 8.2

 Question 1.
ABCD is a quadrilateral in which P, Q, R and S are the mid-points of the sides AB, BC, CD and DA (see figure). AC is a diagonal. Show that: D
(i) SR || AC and SR =  AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.

Solution.
Given : P,Q,R, and S are the mid-points of the sides
AB, BC, CD, and DA, respectively.
AP = BP
BQ = CQ
CR = DR
and AS = DS
To show (i) SR || AC and SR = AC 2
(ii) PQ = SR
(iii) PQRS is a parallelogram.
Proof:
(i) In Δ ADC, we have S is mid-point of AD and R is mid-point of DC.
We know that the line segment joining the mid-points of any two sides of a triangle is parallel to the third side and equal to half of it.
∴ SR || AC …….(i)
SR = AC ………….(ii)

(ii) Similarly, in Δ ABC, we have P is mid-point of AB and Q is the mid-point of BC.
∴ PQ || AC …………..(iii)
PQ =  AC…………..(iv)

(iii) From eqs. (i) and (iii), we get
PQ || SI and from Eq. (v), PQ = SR
Since a pair of opposite sides of a quadrilateral PQRS is equal and parallel.
So, PQRS is a parallelogram.

Question 2.
ABCD is a rhombus 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 rectangle.
Solution.

Given: ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD, and DA, respectively.
To show: Quadrilateral PQRS is a rectangle.
Proof: By mid-point theorem.
In Δ ADC, we have S and R are the mid-points of DA and CD, respectively.
∴  SR || AC
SR =  AC …………..(i)
In Δ ABC, we have P and Q are the mid-points of AB and BC, respectively.
∴ PQ || AC and PQ =  AC …..(ii)
From eqs. (i) and (ii), we get PQ || SR and PQ =SR =  AC
Since, a pair of opposite sides of a quadrilaterals PQRS is equal and parallel. So, PQRS is a parallelogram.

We know that diagonals of a rhombus bisect each other at right angles.
∠COD = ∠EOF = 90°
Now, in ABCD, R and Q are the mid-points of CD and BC, respectively.
RQ || DB   [by mid-point theorem]
⇒ RE || OF
Also,SR || AC [from Eq. (i)]
⇒ FR || OE
So, OERF is a parallelogram.
∴ ∠ERF = ∠EOF = 90°[opposite angles of a parallelogram are equal]
Thus, PQRS is a parallelogram with ∠R =90°
Hence, PQRS is a rectangle.
Hence proved.

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

Given, ABCD is a rectangle.
∴ ∠A=∠B = ∠C =∠D = 90° and AD = BC, AB = CD
Also, given P,Q,R and S are mid-points of AB, BC, CD and DA respectively.
∴ PQ || AC and PQ =  AC ……(i)
AP = BP (Given)
AS= BQ (Given)
∠A= ∠B (Given)
∴ Δ ASP ≅ Δ BQP (By SAS)
∴ SP = PQ  (By CPCT) …(iv)
SD = CQ (Given)
DR = RC (Given)
∠C = ∠D (Given)
Δ  RDS = Δ  RCQ (By SAS)
SR = RQ (By CPCT) …(v)
From Eqs. (iii), (iv) and (v), it is clear that quadrilateral 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 fig.). Show that F is ‘ the mid-point of BC.

Solution.
Given: ABCD is a trapezium in which AB || DC, BD is a diagonal and E is mid-point of AD and a line is drawn through E parallel to AB intersecting BC at F such that EF || AB. To
prove: F is the mid-point of BC.
Proof: Let EF intersects BD at P.
In Δ ABD, we have EP || AB [∵ EF || AB] and E is mid-point of AD.
So, by the converse of mid-point theorem.
We get, P is mid-point of BD.
Similarly, in ABCD, we have PF || CD   [∵ EF || AB and AB || CD]
and P is mid-point of BD.
So, by the converse of mid-point theorem, F is mid-point of BC.

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

Given: ABCD is a parallelogram and E, F are the mid-points of sides AB and CD, respectively.
To prove: Line segments AF and EC trisect the diagonal BD.
Proof : Since, ABCD is a parallelogram.
∴ AB || DC and AB = DC [opposite sides of a parallelogram]
⇒ AF || FC and  AP =  DC
⇒ AF ||  FC and AF = FC
[∴ E and F are the mid-points of AB and CD] Since, a pair of opposite sides of a quadrilaterals AECF is equal and parallel.
So, AECF is a parallelogram.
Then, AF || FC
⇒ AP || FQ and FP || CQ
Since opposite sides of a parallelogram are parallel.
In Δ BAP, E is the mid-point of AB and EQ || AP, so Q is the mid-point of BP.
[by the converse of mid-point theorem]
∴ BQ = PQ     …(i)
Again, in Δ DQC, F is the mid-point of DC and FP || CQ.
So, P is the mid-point of DQ.         [by the converse of mid-point theorem]
∴ QP = DP …(ii)
From eqs. (i) and (ii), we get BQ = PQ = PD
Hence, CE and AF trisect the diagonal BD.
Hence proved.

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

Let ABCD be a quadrilateral and P, Q, R and S be the D mid-points of the sides AB, BC, CD, and DA, respectively.
i.e., AS = DS, AP = BP, BQ = CQ and CR = DR
We have to show that PR and SQ bisect each other,
i.e., SO = OQ and PO = OR
Now, in Δ ADC, S and R are mid-points of AD and CD, respectively.
We know that the line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half of it.
∴ SR || AC and SR = AC    …(i) [by mid-point theorem]
Similarly, in Δ ABC, P and Q are mid-points of AB and BC, respectively.
PQ || AC and PQ = AC [by mid-point theorem]
From Eqs. (i) and (ii), we get
PQ || SP and PQ = SR = AC
So, PQRS is a parallelogram whose diagonals are SQ and also, we know that diagonals of a parallelogram bisect each other. So, SQ and PR bisect each other at O,
i.e. SO = OQ and PO = OR.
[Since a pair of opposite sides of a quadrilaterals PQRS is equal and parallel.]
Hence proved.

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. Show that
(i) D is the mid-point of AC
(ii) MD ⊥ AC
CM = MA = AB
Solution.

Given, ABC is a right angled triangle.
∠C = 90° and M is the mid-point of AB.
Also, DM || BC
(i) In Δ ABC, BC || MD and Mis mid-point of AB.
∴ D is the mid-point of AC.    (By converse of mid-point theorem)
Since, MD || BC and CD is transversal.              .
∠ADM = ∠ACB (Corresponding angles)
But ∠ACB = 90°
∠ADM = 90° => MD ⊥ AC
Now, in AADM and A CDM, we have
DM = MD
AD = CD
∠ADM = ∠MDC
AADM = ACDM
CM = AM
Also, M is mid-point of AB.
AM = BM =  AB
From Eq. (i) and (ii), we get
CM = AM =  AB
Hence proved.