NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1.

- Triangles Class 9 Ex 7.1
- Triangles Class 9 Ex 7.2
- Triangles Class 9 Ex 7.3
- Triangles Class 9 Ex 7.4
- Triangles Class 9 Ex 7.5

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 9 |

Subject |
Maths |

Chapter |
Chapter 7 |

Chapter Name |
Triangles |

Exercise |
Ex 7.1 |

Number of Questions Solved |
8 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1

Question 1.

In a quadrilateral ACBD, AC = AD, and AB bisects ∠A (see figure). Show that Δ ABC ≅ ΔABD. What can you say about and BD?

Solution:

Given : In quadrilateral ACBD.

AC = AD and ∠BAC = ∠DAB

**To prove :** ΔABC = ΔABD

**Proof:** In ΔABC and ΔABD, AC = AD [given]

∠BAC = ∠DAB [given]

and B = AB [common side]

Hence, ΔABC = ΔABD [by SAS congrunce rule]

Then, BC = BD

Thus, BC and BD are equal.

Question 2.

ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see figure). Prove that

(i) Δ ABD ≅ Δ BAC

(ii) BD = AC

(iii) Δ ABD = Δ BAC

Solution:

**Given :** In quadrilateral ABCD.

AD = BC and ∠DAB = ∠CBA

**To prove :
**(i) ΔABD ≅ ΔBAC

(ii) BD= AC

(iii) ∠ABD = ∠BAC

**Proof : (i)**In ΔABD and ΔBAC,

AD-BC [given]

ΔDAB = ΔCBA [given]

and AB = AB [common side]

∴ Δ ABD ≅ Δ BAC [by SAS congruence rule]

**(ii)** From part (i), Δ ABD ≅ ΔBAC

Then, BD = AC [by CPCT]

**(iii)** From part (i), Δ ABD ≅ Δ BAC

Then, ∠ABD = ∠BAC [by CPCT]

Hence proved.

Question 3.

AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.

Solution:

**Given:** AD ⊥ AB and BC ⊥ AB

**To prove:** CD bisects AB.

**Proof:** In Δ AOD and Δ BOC.

AD = BC [given]

∠OAD = ∠OBC [each 90°]

and ∠AOD =∠BOC [vertically opposite angles]

∠AOD = ∠BOC [by AAS congruence rule]

Then,OA = OB [by CPCT]

Thus, CD bisects AB.

Question 4.

l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that

ΔABC = ΔCDA.

Solution:

Given : l || m and p || q

To prove : ΔABC = ΔCDA

Proof : In ΔABC and ΔADC,

∠BAC = ∠ACD [alternative interior angles as p || q]

∠ACB = ∠DAC [alternative interior angles as l || m]

AC = AC [common side]

ΔABC ≅ ΔCDA [by AAS congruence rule]

Question 5.

Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A

(see fig.). Show that

(i) ΔAPB = ΔAQB

(ii) BP = BQ or B is equidistant from the arms of ∠A

Solution:

Consider ΔAPB and ΔAQB, we have

∠APB = ∠AQB =90°

∠PAB = ∠QAB [∵ AB bisects ∠PAQ]

AB = AB [common]

∴ By AAS congruence axiom, we have

ΔAPB ≅ ΔAQB, which proves (i)

=> BP = BQ [by CPCT]

i.e., B is equidistant from the arms of ∠A, which proves (ii).

Question 6.

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

Solution:

In Δ ABC and Δ ADE,

We have AB = AD (Given)

∠BAD = ∠EAC (Given) …(i)

On adding ∠DAC on both sides in Eq. (i)

∠BAD + ∠DAC = ∠EAC + ∠DAC

∠BAC = ∠DAE

and AC – AE [given]

Δ ABC ≅ Δ ADE [by AAS congruence rule]

BC = DE [by CPCT]

Question 7.

AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA ∠DPB (see figure). Show that

(i) Δ DAP ≅ ΔEBP

(ii) AD = BE

Solution:

We have AP = BP [∵ P is the mid-point of AB (Given) ]…(i)

∠EPA = ∠DPB (Given) …. (ii)

∠BAD = ∠ABE (Given) …(iii)

On adding ∠EPD on both sides in Eq. (ii),

we have => ∠EPA + ∠EPD = ∠DPB + ∠EPD

=> ∠DPA = ∠EPB …(iv)

Now, in Δ DAP and Δ EBP, we have …..(iv)

∠DPA = ∠EPB [from eq.(iv)]

∠DAP = ∠EBP (Given)

and AP = BP [from eq.(i)]

ΔDAP ≅ ΔEBP [by ASA congruence axiom]

Hence,AD = BE [by CPCT]

Question 8.

In right triangle ABC, right-angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see fig.). Show that:

(i) Δ AMC = Δ BMD

(ii) ∠DBC is a right angle

(iii) Δ DBC = Δ ACB

(iv) CM=AB

Solution:

Given: Δ ACB in which ZC = 90° and M is the mid-point of AB.

To Prove :

(i) Δ AMC ≅ Δ BMD

(ii) ∠DBC is a right angle

(iii) Δ DBC ≅ ΔACB

(iv) CM = AB

Proof : Consider Δ AMC and Δ BMD,

we have AM – BM

CM = DM

∠AMC = ∠BMD

∴ Δ AMC = Δ BMD

AC = DB

and ∠1–∠2

But ∠1 and ∠2 are alternate angles.

=> BD || CA

Now, BD || CA and BC is transversal.

∴ ∠ACB + ∠CBD = 180° =>90°+ ∠CBD = 180° => ∠CBD =90°

In ΔDBC and ΔACB, we have

CB = BC [common]

DB = AC [using eq.(i)]

∠CBD = ∠BCA [each= 90°]

ΔDBC ≅ = ΔACB [by ASA congruence axiom]

=> DB = AB => DC => AB=CM

=> CM = [∵ CM = DC]

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