NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1

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

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1 are the part of NCERT Solutions for Class 9 Maths. In this post, you will find the NCERT Solutions for Chapter 7 Triangles Ex 7.1.

## 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:

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:

DAB = CBA.

To prove:
(i) Î”ABD  Î”BAC
(ii) BD = AC
(iii)
ABD = BAC
Proof: (i) In Î”ABD and Î”BAC,
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:

AB and BC AB
To prove: CD bisects AB.

Proof: In Î”AOD and Î”BOC.
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
To prove: Î”ABC
Î”CDA
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
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:

DAC on both sides in Eq. (i), we get
BAD + DAC = EAC + DAC
BAC = DAE
and AC = AE                      [given]
Î”ABC
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

Solution:

(i) EPA = DPB                     [Given]         …. (i)
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

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