**NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1****NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2****NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3**

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

**Ex 7.1 Class 9 Maths Question 1.****
**In a quadrilateral ACBD, AC =
AD, and AB bisects ∠A (see figure). Show that Î”ABC ≅ Î”ABD. What can you say about BC and BD?

**Solution:**

**Given:**In quadrilateral ACBD, AC = AD and ∠BAC = ∠DAB

**To prove:** Î”ABC ≅ Î”ABD

**Proof:** In
Î”ABC and Î”ABD, AC = AD [given]

∠BAC = ∠DAB
[given]

and AB = AB
[common side]

Hence, Î”ABC ≅ Î”ABD [by SAS congruence
rule]

Then, BC = BD

Thus, BC and BD are equal.

**Ex 7.1 Class 9 Maths Question 2.****
**ABCD is a quadrilateral in which AD
= BC and ∠DAB = ∠CBA (see figure). Prove that

**(i)**Î”ABD ≅ Î”BAC

**(ii)**BD = AC

**(iii)**∠ABD = ∠BAC

**Solution:**

**Given:**In quadrilateral ABCD, AD = BC and ∠DAB = ∠CBA.

**To prove:**

**(i)** Î”ABD ≅ Î”BAC**
(ii)** BD = AC**
(iii)** ∠ABD = ∠BAC

**Proof:
(i)** In Î”ABD and Î”BAC,

AD = BC
[given]

∠DAB = ∠CBA [given]

and AB = AB
[common side]

∴ Î”ABD ≅ Î”BAC [by SAS congruence rule]

**(ii)** From
part (i), Î”ABD ≅ Î”BAC

Then, BD = AC [by
CPCT]

**(iii)** From
part (i), Î”ABD ≅ Î”BAC

Then, ∠ABD = ∠BAC [by CPCT]

Hence proved.

**Ex 7.1 Class 9 Maths Question 3.
**AD and BC are equal perpendiculars to a line segment
AB (see figure). Show that CD bisects AB.

**Solution:
Given:** AD
= BC, AD

⊥ AB and BC ⊥ AB

**To prove:**CD bisects AB.

**Proof:** In Î”AOD and Î”BOC.

AD = BC [given]

∠OAD = ∠OBC [each 90°]

and ∠AOD = ∠BOC [vertically opposite angles]

Î”AOD ≅ Î”BOC [by
AAS congruence rule]

Then, OA = OB [by CPCT]

Thus, CD bisects AB.

**Ex 7.1 Class 9 Maths Question 4.****
**l and m are two parallel lines
intersected by another pair of parallel lines p and q (see figure). Show that Î”ABC
≅ Î”CDA.

**Solution:**

Given:l || m and p || q

Given:

**To prove:**Î”ABC ≅ Î”CDA

**Proof:**In Î”ABC and Î”ADC,

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

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

AC = AC [common side]

Î”ABC ≅ Î”CDA [by AAS congruence rule]

**Ex 7.1 Class 9 Maths 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 figure). Show that

**(i)**Î”APB ≅ Î”AQB

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

**Solution:**

(i)Consider Î”APB and Î”AQB, we have

(i)

∠APB = ∠AQB = 90°

∠PAB = ∠QAB [∵ AB bisects ∠PAQ]

AB = AB [common]

∴ Î”APB ≅ Î”AQB [by AAS congruence rule]

**(ii)**From part (i), Î”APB ≅ Î”AQB.

Then,
BP = BQ [by
CPCT]

i.e.,
B is equidistant from the arms of ∠A.

**Ex 7.1 Class 9 Maths Question 6.
**In figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC =
DE.

**Solution:**

In Î”ABC and Î”ADE,

AB = AD [Given]

∠BAD = ∠EAC [Given] …(i)

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

∠BAD + ∠DAC = ∠EAC + ∠DAC

∠BAC = ∠DAE

and AC = AE
[given]

Î”ABC ≅ Î”ADE
[by SAS congruence rule]

∴ BC = DE [by CPCT]

**Ex 7.1 Class 9 Maths 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)**Î”DAP ≅ Î”EBP

**(ii)**AD = BE

**Solution:**

**(i)**
∠EPA = ∠DPB [Given] …. (i)

∠BAD = ∠ABE [Given] .…(ii)

On adding ∠EPD on both sides in Eq. (i), we
get

∠EPA + ∠EPD = ∠DPB + ∠EPD

∠DPA = ∠EPB …(iii)

Now, in Î”DAP and Î”EBP, we have

∠DPA = ∠EPB [from eq.(iii)]

∠DAP = ∠EBP [from eq.(ii)]

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

Î”DAP ≅ Î”EBP [by ASA congruence rule]

**(ii)** Since Î”DAP ≅ Î”EBP

AD
= BE
[by CPCT]

**Ex 7.1 Class 9 Maths**** ****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 figure). Show that

**(i)**∆AMC ≅ ∆BMD

**(ii)**∠DBC is a right angle

**(iii)**∆DBC ≅ ∆ACB

**(iv)**CM = ½ AB

**Solution:**

**(i)**In ∆AMC and ∆BMD,

CM = DM [Given]

∠AMC = ∠BMD [Vertically opposite angles]

AM = BM [Since M is the mid-point of AB]

∴ ∆AMC ≅ ∆BMD [By SAS congruence rule]

**(ii)** Since ∆AMC ≅ ∆BMD

⇒ ∠MAC = ∠MBD [By
CPCT]

But they form a pair of alternate interior angles.

∴ AC || DB

Now, BC is a transversal which intersects parallel lines AC and DB,

∴ ∠BCA + ∠DBC = 180° [Co-interior angles]

But ∠BCA = 90° [∆ABC is right angled at
C]

∴ 90° + ∠DBC = 180°

⇒ ∠DBC = 90°

**(iii)** Again, since ∆AMC ≅ ∆BMD

∴ AC = BD [By CPCT]

Now, in ∆DBC and ∆ACB, we have

BD = CA [Proved
above]

∠DBC = ∠ACB [Each 90°]

BC = CB [Common]

∴ ∆DBC ≅ ∆ACB [By SAS congruence rule]

**(iv)**
As ∆DBC ≅ ∆ACB

DC = AB [By CPCT]

But DM = CM [Given]

∴ CM = ½ DC = ½ AB

⇒ CM = ½ AB

**Related Links:**

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**NCERT Solutions for Maths Class 11**