ML Aggarwal Solution Class 9 Chapter 10 Triangles Exercise 10.2

 Exercise 10.2

Question 1

In triangles ABC and PQR, ∠A= ∠Q and ∠B = ∠R. Which side of ΔPQR should be equal to side AB of ΔABC so that the two triangles are congruent? Give reason for your answer.

Sol :

In ΔABC and ΔPQR

⇒∠A=∠Q

⇒∠B=∠R

⇒AB=QP

∵Two Δs are congruent of their corresponding two angles and included sides are equal.

Question 2

In triangles ABC and PQR, ∠A = ∠Q and ∠B = ∠R. Which side of ΔPQR should be equal to side BC of AABC so that the two triangles are congruent? Give reason for your answer.

Sol :

In ΔABC and ΔPQR

and their included sides AB and QR will be equal for their congruency

∴BC=PR (c.p.c.t)

Question 3

“If two angles and a side of one triangle are equal to two angles and a side of another triangle, then the two triangles must be congruent”. Is the statement true? Why?

Sol :

The given statement can be true only if the corresponding (included) sides are equal otherwise not.

Question 4

In the given figure, AD is median of ∆ABC, BM and CN are perpendiculars drawn from B and C respectively on AD and AD produced. Prove that BM = CN.

Sol :

Given : In ∆ABC, AD is median BM and CN are perpendicular to AD from B and C respectively

To prove : In ΔBMD and ΔCND

Proof : In ΔBMD and ΔCND 

⇒BD=CD (AD is median)

⇒∠M=∠N (each 90°)

⇒∠BDM=∠CDN (Vertically opposite angles)

∴ΔBMD=ΔCND (AAS axiom)

∴BM=CN (c.p.c.t)

Question 5

In the given figure, BM and DN are perpendiculars to the line segment AC. If BM = DN, prove that AC bisects BD.

Sol :

Given : In the figure , BM and DN are perpendicular to AC

⇒BM=DN

To prove : AC bisects BD i.e. BE=ED

Construction : Join BD which intersects AC at E

Proof : In ΔBEM and ΔDEN

⇒BM=DN (Given)

⇒∠M=∠N (each 90°)

⇒∠DEN=∠BEM (vertically opposite angles)

∴ΔBEM≅ΔDEN (AAS axiom)

∴BE=ED

⇒AC bisects BD

Question 6

In the given figure, l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that ∆ABC ≅ ∆CDA.

Sol :

In the given figure, two lines l and m are parallel to each other and lines p and q are also a pair of parallel lines intersecting each other at A,B,C and D . AC is joined.

To prove : ΔABC≅ΔCDA

Proof : In ΔABC and ΔCDA

⇒AC=AC (Common)

⇒∠ACB=∠CAD (Alternate angles)

⇒∠BAC=∠ACD (Alternate angles)

∴ΔABC≅ΔCDA (ASA axiom)

Question 7

In the given figure, two lines AB and CD intersect each other at the point O such that BC || DA and BC = DA. Show that O is the mid-point of both the line segments AB and CD.

Sol :

Given : In the given figure, lines AB and CD intersect each other at O such that BC||AD and BC=DA

To prove : O is the mid-point of AB and CD

Proof : ΔAOD and ΔBOC

⇒AD=BC (Given)

⇒∠OAD=∠OBC (Alternate angles)

⇒∠ODA=∠OCB (Alternate angles)

∴ΔAOD≅ΔBOC (SAS axiom)

∴OA=OB and OD=OC

∴ O is the mid-point of AB and CD

Question 8

In the given figure, ∠BCD = ∠ADC and ∠BCA = ∠ADB. Show that

(i) ∆ACD ≅ ∆BDC

(ii) BC = AD

(iii) ∠A = ∠B.

Sol :

Given : In the figure,

⇒∠BCD=∠ADC

⇒∠BCA=∠ADB

To prove : 

(i) ∆ACD≅∆BDC

(ii) BC=AD

(iii) ∠A=∠B

Proof : ∵∠BCA=∠ADB

⇒and ∠BCD=∠ADC

Adding we get,

⇒∠BCA+∠BCD=∠ADB+∠ADC

⇒∠ACD=∠BDC

Now in ΔACD and ΔBDC

⇒CD=CD (Common)

⇒∠ACD=∠BDC (Proved)

⇒∠ADC=∠BCD (Given)

(i) ∴ΔACD≅ΔBDC (ASA axiom)

∴AD=BC (c.p.c.t)

⇒∠A=∠B (c.p.c.t)

Question 9

In the given figure, ∠ABC = ∠ACB, D and E are points on the sides AC and AB respectively such that BE = CD. Prove that

(i) ∆EBC ≅ ∆DCB

(ii) ∆OEB ≅ ∆ODC

(iii) OB = OC.

Sol :

Given : In the given figure;

⇒∠ABC=∠ACB

⇒D and E are the points on AC and AB such

To prove : ∆EBC ≅ ∆DCB

(ii) ∆OEB ≅ ∆ODC

(iii) OB = OC.

Proof : In ∆ABC

∵∠ABC=∠ACB

∴AC=AB (Sides opposite to equal angles)

In ∆EBC and ∆DCB

⇒EB=DC (Given)

⇒BC=BC (Common)

⇒∠CBD=∠DCB (∵∠ABC=∠ACB)

(i) ∵ ∆EBC ≅ ∆DCB (SAS axiom)

⇒∠ECB=∠DBC (c.p.c.t)

Now in ∆OEB and ∆ODC

⇒BE=CD (Given)

⇒∠EBO=∠DCO

{∵∠ABC-∠DBC=∠ACB-∠OCB}

(ii) ∴∆OEB ≅ ∆ODC (AAS axiom)

(iii) ∴OB=OC (c.p.c.t)

Question 10

ABC is an isosceles triangle with AB=AC. Draw AP ⊥ BC to show that ∠B = ∠C.

Sol :

Given :

AB=AC

AP⏊BC

To prove : ∠B=∠C

Proof : In right ΔAPB and ΔAPC

Side AP=AP (Common)

Hyp.AB=AC (Given)

∴ΔAPB≅ΔAPC (RHS axiom)

∴∠B=∠C (c.p.c.t)

Question 11

In the given figure, BA ⊥ AC, DE⊥ DF such that BA = DE and BF = EC.

Sol :

Given : In the given figure ,

BA⊥AC, DE⊥DF

BA=DE , BF=EC

To prove : ΔABC≅ΔDEF

Proof : ∵BF=CE

Adding FC both sides

⇒BF+FC=FC+CE

⇒BC=EF

Now in right ΔABC and ΔDEF 

Side AB=DE (Given)

Hypo. BC=EF (Proved)

∴ΔABC≅ΔDEF

Question 12

ABCD is a rectangle. X and Y are points on sides AD and BC respectively such that AY = BX. Prove that BY = AX and ∠BAY = ∠ABX.

Sol :

Given : In rectangle ABCD,X and Y are points on the sides AD and BC respectively such that AY=BX

To prove : BY=AX and ∠BAY=∠ABX

Proof : In ΔABX and ΔABY

⇒AB=AB (Common)

⇒∠A=∠B (Each 90°)

⇒BX=AY

∴ΔABX≅ΔABY (SAS axiom)

⇒AX=BY (c.p.c.t)

⇒or BY=AX

and ∠AXB=∠BYA (c.p.c.t)

Question 13

(a) In the figure (1) given below, QX, RX are bisectors of angles PQR and PRQ respectively of A PQR. If XS⊥ QR and XT ⊥ PQ, prove that

(i) ∆XTQ ≅ ∆XSQ

(ii) PX bisects the angle P.

(b) In the figure (2) given below, AB || DC and ∠C = ∠D. Prove that

(i) AD = BC

(ii) AC = BD.

(c) In the figure (3) given below, BA||DF and CA||EG and BD = EC . Prove that, .

(i) BG = DF

(ii) EG = CF.

Sol :

(a) Given : In ΔPQR, QX is the bisector of ∠PRQ.

⇒XS⊥QR and XT⊥PQ

To prove : 

(i) ∆XTQ≅∆XSQ

(ii) PX bisects the angle P

Construction : Draw XZ⊥PR

Proof : In ∆XTQ and ∆XSQ 

⇒∠QTX=∠QSX  (each 90°)

∴[XT⊥QR (given) and XT⊥PQ (given)]

⇒∠TQX=∠SQX (QX is bisector of ∠PQR)

⇒QX=QX (common)

∴ΔXTQ≅ΔXSQ 

[By A.A.S axiom of congruency]

(ii) XT=XS (c.p.c.t) ..(1)

Now, in ΔXSR and ΔXZR

⇒∠XSR=∠XZR (each 90°)

[∴∠XSR=90° (given) ∠XZR=90° by construction]

⇒∠SRX=∠ZRX [RX is the bisector of ∠PRQ]

⇒RX=RX (common)

∴ΔXSR≅ΔXZR (By A.A.S axiom of congruency)

∴XS=XZ (c.p.c.t) ...(2)

From (1) and (2) , we get

⇒XT=XZ ..(3)

In ΔXTP and ΔXZP,

⇒∠XTP=∠XZP=90° $\left[\begin{array}{c}\angle \mathrm{XTP}=90^{\circ}(\text { given }) \\ \angle \mathrm{XTP}=90^{\circ}(\text { construction })\end{array}\right]$

⇒(hyp) XP=(hyp)XP (common)

XT=XZ [From (3)]

∴ΔXTP≅ΔXZP [By R.H.S axiom of congruency]

∴∠XPT=∠XPZ (c.p.c.t)

∴PX bisects the angle P (Q.E.D)

(b) In following figure 

Given : AB||DC and ∠C=∠D

To prove : (i)AD=BC

(ii) AC=BD

<Diagram to be added>

Construction : Draw AE⊥CD, BF⊥CD and Join A to C and B to D

Proof : (i) In ΔAED and ΔBCF

⇒∠AED=∠BFC  (each 90°)

[By construction AE⊥CD and BF⊥CD]

⇒∠D=∠C (given)

⇒AE=BF

[Distance between parallel lines are same]

∴ΔAED≅ΔBCF (By A.A.S axiom of congruency)

⇒AD=BC (c.p.c.t)..(1)

(ii) In ΔACD and ΔBCD

⇒∠D=∠C (Given)

⇒DC=DC

⇒AD=BC [From (1)]

∴ΔACD≅ΔBCD (By S.A.S axiom of congruency)

∴AC=BD (c.p.c.t) (Q.E.D)

(c) In following figure

Given : BA||DF and CA|||EG and BD=EC 

To prove : (i) BG=DF

(ii) EG=CF

<Diagram to be added>

Proof : (i) In ΔBEG and ΔDCF

∠B=∠D (∵ BA||DF , corresponding angles equal)

∠E=∠C (∵ CA||EG, corresponding angles equal)

and BE=BC-EC=BC-BD=DC

i.e. BE=DC

∴ΔBEG≅ΔDCF (By A.S.A axiom of congruency)

∴BG=DF (c.p.c.t)

(iii) EG=CF (c.p.c.t) (Q.E.D)

Question 14

In each of the following diagrams, find the values of x and y.

Sol :

In ΔABC and ΔCDE

BC=CD (Given)

⇒∠BAC=∠CED (Given)

⇒∠ACB=∠DCE (Vertically opposite angles)

∴BC=CD (c.p.c.t)

⇒3x-7=32

⇒3x=32+7

⇒3x=39

⇒$x=\frac{39}{3}$

⇒x=13

Also , AB=DE (c.p.c.t)

⇒25=2y+33

⇒2y+3=25

⇒2y=25-3

⇒2y=22

⇒y=11

Hence ,x=13 and y=11

(ii) <Diagram to be added>

In ΔABC and ΔAED

⇒AC=AD (given)

⇒∠BAC=∠DAE (given)

⇒∠ACB=∠ADE (given)

∴ΔABC≅ΔAED (By A.A.S axiom of congruency)

∴AB=AE (c.p.c.t)

⇒2x+4=3y+8

⇒2x-3y=8-4

⇒2x-3y=4..(1)

Also, BC=DE (c.p.c.t)

⇒x=2y

⇒x-2y=0 ..(2)

Multiplying equation (2) by 2 and subtracting equation (1), we get

Subtracting

$\begin{array}{l}2 x-4 y=0 \\2 x-3 y=4 \\-\quad+\quad- \\\hline-y=-4\end{array}$

⇒y=4

Substituting the value of y in equation (2) we get

⇒x-2×4=0

⇒x-8=0

⇒x=8

Hence ,x=8 , y=4