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. Here you can find the NCERT Solutions for 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 ΔABC ≅ 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 AB = AB                             [common side]
Hence, ΔABC ≅ ΔABD                            [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) ΔABD ≅ ΔBAC
(ii) BD = AC
(iii) ∠ABD = ∠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]
∠DAB = ∠CBA                          [given]
and AB = AB                           [common side]
∴ ΔABD ≅ ΔBAC                      [by SAS congruence rule]

(ii) From part (i), ΔABD ≅ ΔBAC
Then, BD = AC [by CPCT]

(iii) From part (i), ΔABD ≅ ΔBAC
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 ΔAOD and ΔBOC.
AD = BC                             [given]
∠OAD = ∠OBC                 [each 90°]
and ∠AOD = ∠BOC          [vertically opposite angles]
ΔAOD ≅ ΔBOC                  [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 ΔABC ≅ ΔCDA.

Solution:
Given: l || m and p || q
To prove: ΔABC ≅ ΔCDA
Proof: In ΔABC and ΔADC,
∠BAC = ∠ACD                       [alternate interior angles as p || q]
∠ACB = ∠DAC                       [alternative interior angles as l || m]
AC = AC                                 [common side]
ΔABC ≅ ΔCDA                      [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) ΔAPB ≅ ΔAQB
(ii) BP = BQ or B is equidistant from the arms of ∠A

Solution:
(i) Consider ΔAPB and ΔAQB, we have
∠APB = ∠AQB = 90°
∠PAB = ∠QAB                    [∵ AB bisects ∠PAQ]
AB = AB                           [common]
∴ ΔAPB ≅ ΔAQB                [by AAS congruence rule]
(ii) From part (i), ΔAPB ≅ ΔAQB.

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 ΔABC and ΔADE,
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]
ΔABC ≅ ΔADE                   [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) ΔDAP ≅ ΔEBP
(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 ΔDAP and ΔEBP, we have
∠DPA = ∠EPB                         [from eq.(iii)]
∠DAP = ∠EBP                         [from eq.(ii)]

AP = BP [∵ P is the mid-point of AB (Given)]
ΔDAP ≅ ΔEBP [by ASA congruence rule]
(ii) Since ΔDAP ≅ ΔEBP

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

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NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1

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