Well-structured OP Malhotra Class 9 Maths Solutions Chapter 8 Triangles Ex 8(A) facilitate a deeper understanding of mathematical principles.

## S Chand Class 9 ICSE Maths Solutions Chapter 8 Triangles Ex 8(A)

Question 1.

In the figure, AB and CD bisect each other at K. Prove that AC = BD.

Solution:

Given: Two line segments AB and CD bisect each other at K. AC and BD are joined.

To prove : AC = BD

Proof: In ∆AKC and ∆BKD

AK = BK (given)

CK = DK (given)

∠AKC = ∠BKD (vertically opposite angles)

∆AKC ≅ ∆BKD (SAS axiom)

AC = BD (c.p.c.t.)

Hence proved.

Question 2.

In the figure, the sides BA and CA have been produced such that BA = AD and CA = AE. Prove that DE || BC.

Solution:

Given : In ∆ABC, sides BA and CA are produced such that BA = AD and CA = AE. ED is joined

To prove : DE || BC

Proof: In ∆BAC and ∆DAE

BA = DA (given)

CA = EA (given)

∠BAC = ∠DAE (vertically opposite angles)

∴ ∆BAC = ∆DAE (SAS axiom)

∴ ∠ABC = ∠ADE (c.p.c.t.)

But these are alternate angles

∴ DE || BC

Hence proved.

Question 3.

In the figure, ABCD is a rectangle; P is the; mid-point of AB, Q and R are points in AD and BC respectively such that AQ = BR. Prove that PQ = PR.

Solution:

Given : ABCD is a rectangle. P is the mid point of AB. Q and R are points in AD and BC respectively such that AQ = BR PO and PR are joined

To prove : PQ = PR

Proof: In ∆PAQ and ∆PBR

PA = PB (∵ P is mid point of AB)

∠A = ∠B (each 90°)

AQ = BR (given)

∴ ∆PAQ ≅ ∆PBR (SAS axiom)

PQ = PR (c.p.c.t.)

Hence proved.

Question 4.

In the figure, OA = OB, OC = OD, ∠AOB = ∠COD, prove that AC = BD.

Solution:

Given : In the figure OA = OB

OC = OD, ∠AOB = ∠COD

To prove : AC = BD

Proof: ∵ ∠AOB = ∠COD

∴ ∠AOC + ∠COB = ∠DOB + ∠COB

⇒ ∠AOC = ∠BOD

In ∆OAC and ∆ODB

OA = OB (given)

OC = OD (given)

and ∠AOC = ∠BOD (proved)

∴ ∆OAC ≅ ∆ODB (SAS axiom)

∴ AC = BD (c.p.c.t)

Hence Proved.

Question 5.

In figure, prove that ∆s OAM and OBN are congruent and hence prove that AM = BN.

Solution:

Given : In the figure

OA = OB

AM ⊥ XY and BN ⊥ XY

To prove :

(i) ∆OAM s ∆OBN

(ii) AM ≅ BN

Proof:

(i) In ∆OAM and ∆OBN

OA = OB (given)

∠M = ∠N (each 90°)

and ∠AOM = ∠BON

(vertically opposite angles)

∴ ∆OAM ≅ ∆OBN (AAS axiom)

(ii) Hence AM = BN (c.p.c.t.)

Question 6.

In Figure, ∠XYZ is bisected by YP. L is any point in YP and MLN is perpendicular to YP. Prove that LM = LN.

Solution:

Given : In the figure,

YP is the bisector of ∠XYZ

L is any point on YB and MLN is perpendicular to YP

To prove : LM = LN

Proof: In ∠YML and ∆YNL

YL = YL (common)

∠YLM = ∠YLN (each 90°)

∠MYL = ∠NYL (∵ YP is the bisector of ∠XYZ)

∴ ∆YML ≅ ∆YNL (ASA axiom)

∴ LM = LN (c.p.c.t.)

Hence proved.

Question 7.

In figure, X is any point within a square ABCD. On AX a square AXYZ is described. Prove that BX = DZ.

Solution:

Given : In the figure,

ABCD and AXYZ are squares where X is any point within the square ABCD

BX and DZ are joined

To prove : BX = DZ Proof: In ∆ABX and ∆ADZ,

AB = AD (sides of the square ABCD)

AX = AZ (sides of the square AXYZ)

∠B AX = ∠DAZ (each = 90° – ∠D AX)

∴ ∆ABX = ∆ADZ (SAS axiom)

∴ BX = DZ (c.p.c.t.)

Hence proved.

Question 8.

In figure, prove that AP bisects ∠BAC.

Solution:

Given : In the figure, P is any point inside the ∠BAC

PM ⊥ AB and PN ⊥ A are drawn and PM = PN

To prove : AP is the bisector of ∠BAC

Proof : In right ∆PAM and ∆PAN

Side PM = PN (given)

Hyp PA = PA (common)

∴ ∆PAM ≅ ∆PAN (RHS axiom)

∴ ∠PAM – ∠PAN (c.p.c.t.)

∴ PA is the bisector of ∠B AC

Hence proved.

Question 9.

In figure, if PR – PS, PT = PQ, ∠PTS = ∠PQR = 90°, prove that RQ – ST and RT = SQ.

Solution:

Given : In the figure, PR = PS, PT = PQ, ∠PQR = 90° and ∠PTS = 90°

To prove :

(i) RQ = ST

(ii) RT = SQ

Proof:

(i) In right ∆PQR and ∆PTS

Side PQ = PT (given)

Hyp. PR = PS (given)

∴ ∆PQR ≅ ∆PTS (RHS axiom)

∴ RQ = ST (c.p.c.t)

(ii) ∵ PR = PS

and PT = PQ

Subtracting we get,

PR – PT = PS – PQ

⇒ RT = SQ

Hence proved.

Question 10.

In figure, PQRS is a square. Arc AB is drawn with centre P and any radius less than PR, cutting SR at A and RQ at B. Prove that AS = BQ.

Solution:

Given : In square PQRS an arc AB is drawn with centre P and a radius less then PR which intersects SR at A and RQ at B

To prove : AS = BQ

Construction : Join PA and PB

Proof : In right ∆PQB and ∆PSA

Side PQ = PS (sides of a square)

Hyp. PB = PA (radii of the same arc AB)

∴ ∆PQB = ∆PSA (RHS axiom)

∴ BQ = AS (c.p.c.t.)

or AS = BQ

Hence proved.

Question 11.

In figure, if LM = MN, QM = MR and ∠MLQ = ∠MNR = 90°, prove that PQ = PR.

Solution:

Given : In the figure, ∆PQR in which

LM = MN, QM = MR and

∠MLQ = ∠MNR = 90°

To prove : PQ = PR

Proof: In right ∆QLM and ∆RNM

Side LM = MN (given)

Hyp. QM = MR

∴ ∆QLM = ∆RNM (RHS axiom)

∴ ∠Q = ∠R (c.p.c.t.)

Now in APQR

∵ ∠Q = ∠R (proved)

∴ PR = PQ (sides opposite equal angles)

or PQ = PR

Hence proved.

Question 12.

In figure, ∆ABC is right-angled at B. ∆CDE and ∆CGF are squares. Prove that

(i) ∆BCD ≅ ∆ACG;

(ii) AG = BD

Solution:

Given : ∆ABC is a right angled, right angle at B. ACDE and BCGF are squares on the sides AC and BC respectively. AG and BD are joined

To prove :

(i) ∆BCD = ∆ACG

(ii) AG = BD

Proof: ∵ ∠ACD = ∠BCG (each 90°)

Adding ∠ACB to both sides,

∠ACD + ∠ACB = ∠ACB + ∠BCG

⇒ ∠BDC = ∠ACG

Now in ∆BCD and ∆ACG

BC = CG (sides of the same square)

∠BCD = ∠ACG (proved)

CD = AC (side of the same square)

∴ ∆BCD ≅ ∆ACG (SAS axiom)

∴ BD = AG (c.p.c.t.)

or AG = BD

Hence proved.

Question 13.

Prove that the medians of an equilateral triangle are equal.

Solution:

Given : ∆ABC is an equilateral triangle in which AD, BE and CF are its medians.

To prove : AD = BE = CF

Proof: In ∆EBC and ∆FBC

BC = BC (common)

EC = FB

(Half of equal sides of equilateral triangle)

∠C = ∠B (each 90°)

∴ ∆EBC ≅ ∆FBC (SAS axiom)

∴ BE = CF … (i)

Similarly we can prove that

∆EBC ≅ ∆ADC

∴ BE = AD …. (ii)

From (i) and (ii),

AD = BE = CF

Hence proved.

Question 14.

In figure, AC = DE, ∠ACB =∠EDF and BD = CF. Prove that AB = EF.

Solution:

Given : In the figure,

AC = DE, ∠ACB =∠EDF

and BD = FC

To prove : AB = EF

Proof:

∵ BD = CF

Adding DC to both sides,

BD + DC = DC + CF

⇒ BC = DF

Now in ∆ABC and ∆DEF

AC = DE (given)

BC = DF (proved)

∠ACB = ∠EDF (given)

∴ ∆ABC ≅ ∆DEF (SAS axiom)

∴ AB = EF (c.p.c.t.)

Hence proved.

Question 15.

In figure, AC = AE, AB = AD and ∠BAD =∠EAC. Prove that BC = DE.

Solution:

In the figure,

AC = AE, AB = AD and ∠BAD = ∠EAC

To prove : BC = DE

Construction : Join DE

Proof:

∵ ∠BAD = ∠EAC

∴ ∠BAD + ∠DAC = ∠DAC + ∠EAC

In ∆ABC and ∆ADE

AB = AD (given)

AC = AE (given)

∠BAC = ∠DAE (proved)

∴ ∆ABC ≅ ∆ADE (SAS axiom)

∴ BC = DE (c.p.c.t.)

Hence proved.

Question 16.

In figure, PS is a median and QL and RM are perpendiculars drawn from Q and R respectively on PS and PS produced. Prove that QL = RM.

Solution:

In ∆PQR, PS is the median of side QR

QL and RM are perpendiculars on the median

To prove : QL = RM

Proof:

In ∆QLS and ∆RMS

QS = SR (S is mid point of QR)

∠QSL = ∠RSM (vertically opposite angles)

∠QLS = ∠RMS (each 90°)

∆QLS ≅ ∆RMS (AAS axiom)

∴ QL = RM (c.p.c.t.)

Hence proved.

Question 17.

ABCD is a parallelogram. The sides AB, AD are produced to E, F so that AB = BE and AD = DF. Prove that the triangles BEC and DCF are congruent. (SC)

Solution:

Given : ABCD is a parallelogram, sides AB and AD are produced to E and F respectively such that AB = BE and AD = DF

To prove : ∆BEC ≅ ∆DCF

Construction : Join EC and FC

Proof: In ||gm ABCD,

∠DAB = ∠FDC = ∠CBE (corresponding angles)

Now in ABEC and ∆DCF

BC = DF (each = AD)

BE = DC (each = AB)

and ∠CBE = ∠FDC (proved)

∴ ∆BEC ≅ ∆DCF (SAS axiom)

Hence proved.

Question 18.

In figure, QX and RX are the bisectors of the angles Q and R respectively of the triangle PQR. If XS ⊥ QR and XT ⊥ PQ, prove that,

(i) ∆XTQ ≅ ∆XSQ;

(ii) PX bisects the angle P.

Solution:

Given : In the figure, QX and RX are the bisectors of the angles Q and R respectively of the triangle PQR and XS ⊥ QR and XT ⊥ PQ

To prove :

(i) ∆XTZ ≅ ∆XSQ

(ii) PX bisects the angle P

Construction : Join PX and draw XV ⊥ PR

Proof:

In ∆XQS and ∆XTQ,

XQ = XQ (common)

∠S = ∠T (each 90°)

∠XQS = ∠XQT (∵ XQ is bisector of ∠Q)

∴ ∆XQS ≅ ∆XTQ (AAS)

or ∆XTQ = ∆XSQ

∴ XS = XT (c.p.c.t.) … (i)

Similarly we can prove that

∆XSR ≅ ∆XVR

∴ XS = XV … (ii)

From (i) and (ii) XT = XV

Now in ∆ right XTP and ∆XYP

Hyp. XP = XP (common)

Side XT = XV (proved)

∴ ∆XTP ≅ ∆XYP (RHS axiom)

∴ ∠XPT = ∠XPV (c.p.c.t.)

∴ PX bisects ∠P

Hence proved.

Question 19.

Find the values ofx and y in the figures given below, using congruency of triangles.

Solution:

(i) In the figure (i) AB || DE and BC = CD

In ∆ABC and ∆DEC,

BC = CD (given)

∠ACB = ∠ECD (vertically opposite angles)

∠B = ∠D (alternate angles)

∴ ∆ABC ≅ ∆DEC (ASA axiom)

∴ AB = DE (c.p.c.t.)

⇒ 2x – 4 = 14 ⇒ 2x = 14 + 4 = 18

⇒ x = \(\frac { 18 }{ 2 }\) = 9

∴ x = 9

and AC = CE (c.p.c.t.)

⇒ 3y + 5 = 20 ⇒ 3y = 20 – 5 = 15

⇒ y = \(\frac { 15 }{ 3 }\) = 5

Hence x = 9 and y = 5

(ii) In figure (ii)

AB = AD, BC = DC

∠BAC = (y – 6)°, ∠BCA = 63°

∠CAD = 30° and ∠ACD = (2x + 7)°

Now in ∆ABC and ∆ADC

AC = AC (common)

AB = AD (given)

BC = DC (given)

∴ ∆ABC ≅ ∆ADC (SSS axiom)

∴ ∠BAC = ∠CAD (c.p.c.t)

∴ ∠BAC = 30° ⇒ y – 6° = 30°

⇒ y = 30° + 6° = 36°

and ∠BCA = ∠ACD

63° = 2x + 7° 2x = 63 – 7 = 56°

⇒ x = \(\frac { 56° }{ 2 }\) = 28°

Hence x = 28° and y = 36°