# ML Aggarwal Triangles Exe-10.2 Class 9 ICSE Maths Solutions

ML Aggarwal Triangles Exe-10.2 Class 9 ICSE Maths Solutions Ch-10. Step by Step Answer of Exercise-10.2 of Triangles of ML Aggarwal for ICSE Class 9th Mathematics Questions. Visit official website CISCE for detail information about ICSE Board Class-9.

## ML Aggarwal Triangles Exe-10.2 Class 9 ICSE Maths Solutions

 Board ICSE Subject Maths Class 9th Chapter-10 Triangles Topics Solution of Exe-10.2 Questions Academic Session 2024-2025

### Solution of Exe-10.2 Questions

ML Aggarwal Triangles Class 9 ICSE Maths Solutions Ch-10

#### Question 1. In triangles ABC and PQR, ∠A= ∠Q and ∠B = ∠R. Which side of APQR should be equal to side AB of AABC so that the two triangles are congruent? Give reason for your answer.

Answer: In triangle ABC and triangle PQR

∠A = ∠Q

∠B = ∠R

AB = QP

Because triangles 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 APQR should be equal to side BC of AABC so that the two triangles are congruent? Give reason for your answer.

∠A = ∠Q

∠B = ∠R

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

Therefore, BC = PR by corresponding parts of congruent triangles.

#### 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?

Answer:  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.

Answer: Given in ∆ABC, AD is median BM and CN are perpendicular to AD form B and C respectively.

To prove:

BM = CN

Proof:

In ∆BMD and ∆CND

BD = CD (because AD is median)

∠M = ∠N

∠BDM = ∠CDN (vertically opposite angles)

∆BMD ≅ ∆CND (AAS axiom)

Therefore, BM = CN.

#### Question 5. In the given figure, BM and DN are perpendiculars to the line segment AC. If BM = DN, prove that AC bisects BD.

Answer:  Given in figure BM and DN are perpendicular to AC

BM = DN

To prove:

AC bisects BD that is BE = ED

Construction:

Join BD which intersects Ac at E

Proof:

In ∆BEM and ∆DEN

BM = DN

∠M = ∠N (given)

∠DEN = ∠BEM (vertically opposite angles)

∆BEM ≅ ∆DEN

BE = ED

Which implies 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.

Answer: 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: ∆ABC ≅ ∆CDA

Proof:

In ∆ABC and ∆CDA

AC = AC (common)

∠BAC = ∠ACD (alternate angles)

∆ABC ≅ ∆ DCA (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.

Answer: 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 midpoint of Ab and Cd

Proof:

Consider ∆AOD and ∆BOC

∠ODA = ∠OCB (alternate angles)

∆AOD ≅ ∆BOC (SAS axiom)

Therefore, OA = OB and OD = OC

Therefore O is the midpoint of AB and CD.

#### Question 8. In the given figure, ∠BCD = ∠ADC and ∠BCA = ∠ADB. Show that

(i) ∆ACD ≅ ∆BDC
(iii) ∠A = ∠B.

∠1= ∠2 and ∠3 = ∠4

∠1+∠3 = ∠2+∠4

∠ACD = ∠BDC

In ∆ACD and ∆BDC

CD = CD (COMMON)

∠ACD = ∠BDC [from (i)]

∆ACD  ≅  ∆BDC (ASA rule)

AD = BC and ∠A = ∠B (CPCT)

#### 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.

Answer :  given:-. ∠ABC=∠ACB. , BE=CD

TO prove:-

(i) triangle EBC  triangle DCB

(ii) triangle OEB  triangle ODC

(iii) OB=OC

proof :-

in tri(EBC) and tri(DCB)

∠EBC=∠DCB. ( given ,∠ABC=∠ACB)

BE = CD. (given)

BC= BC. (common)

so , △EBC≡ △DCB. ( by SSS property).

now subtract common (OBC) from both sides

△EBC – (OBC)≡ △DCB -(OBC)

△OEB  △ ODC.

OB = OC ( by C.P.C.T. )

Hence proved

#### Question 10. ABC is an isosceles triangle with AB=AC. Draw AP ⊥ BC to show that ∠B = ∠C.

Answer:  ABC is an isosceles triangle with
AB = AC.
AP ⊥ BC
To Prove: ∠B = ∠C

#### Question 11. In the given figure, BA ⊥ AC, DE⊥ DF such that BA = DE and BF = EC.

Answer : AC = EF, we have to prove ΔABC = ΔDEF

Here BA = DE (given) …(i)

BA ⊥ AC and DE ⊥ FE

∠BAC = ∠DEF = 90° (each) …(ii)

Also BF = CD (given) …(iii)

BF + FC = CD + FC

⇒ BC = FD …(iv)

From (i), (ii) and (iv), we get

ΔABC = ΔDEF (by RHS congruency property)

⇒ AC = EF (by c.p.c.t)

#### 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.

Given that ABCD is a square, X and Y are points on the sides AD and BC respectively.

AY = BX

We have to prove:

BY = AX and ∠BAY = ∠ABX

Join B and X, A and Y

Since, ABCD is a square

∠DAB = ∠CBA = 90°

∠XAB = ∠YBA = 90°………………(i)

Now, consider ΔXAB and ΔYBA

∠XAB = ∠YBA = 90°[From (i)]

BX = AY (Given)

AB = BA (Common side)

So, by RHS congruence rule, we have

ΔXAB ≅ ΔYBA

BY = AX and ∠BAY = ∠ABX(By c.p.c.t)

Hence, proved

#### 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.

(ii) AC = BD.

(i) BG = D
(ii) EG = CF.