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.**

**Answer: **In ∆ABC and ∆PQR

∠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)

∠ACB = ∠CAD (alternate angles)

∠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

AD = BC (given)

∠OAD = ∠OBC (alternate angles)

∠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

(ii) BC = AD

(iii) ∠A = ∠B.

**Answer : **

∠1= ∠2 and ∠3 = ∠4

∠1+∠3 = ∠2+∠4

∠ACD = ∠BDC

In ∆ACD and ∆BDC

∠ADC = ∠BCD (given)

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)

Adding FC in equation (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.**

**Answer :**

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.

**(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 II EG and BD = EC . Prove that, .**

(i) BG = D

(ii) EG = CF.

**Answer :**

**Question 14. ****In each of the following diagrams, find the values of x and y.**

**Answer :**

— : End of ML Aggarwal Triangles Exe-10.2 Class 9 ICSE Maths Solutions :–

Return to :- ** ML Aggarawal Maths Solutions for ICSE Class-9**

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