7. Triangles Mathematics Exercise - 7.1 class 9 Maths in English - CBSE Study
NCERT Solutions for Class 9 Mathematics are carefully prepared according to the latest CBSE syllabus and NCERT textbooks to help students understand every concept clearly. These solutions cover all important 7. Triangles with detailed explanations and step-by-step answers for better exam preparation. Each Exercise 7.1 is explained in simple language so that students can easily grasp the fundamentals and improve their academic performance. The study material is designed to support daily homework, revision practice, and final exam preparation for Class 9 students. With accurate answers, concept clarity, and structured content, these NCERT solutions help learners build confidence and score higher marks in their examinations. Whether you are revising a specific topic or preparing an entire chapter, this resource provides reliable and syllabus-based guidance for complete success in Mathematics.
Class 9 English Medium Mathematics All Chapters:
7. Triangles
1. Exercise 7.1
Chapter 7. Triangles
Exercise 7.1
Q1. In quadrilateral ACBD, AC = AD and AB bisects ∠A (see Fig. 7.16). Show that Δ ABC ≅ Δ ABD.
Solution:
Given: AC = AD and AB bisects ∠A
To prove: Δ ABC ≅ Δ ABD.
Proof: In Δ ABC and Δ ABD.
AC = AD [given]
∠CAB = ∠BAD [AB bisect ∠A]
AB = AB [Common]
By SAS Congruence Criterion Rule
Δ ABC ≅ Δ ABD
BC = BD [By CPCT] Proved
Q2. ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see Fig.7.17). Prove that
(i) Δ ABD ≅ Δ BAC
(ii) BD = AC
(iii) ∠ ABD = ∠ BAC
Solution:
Given: ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA
To prove:
(i) Δ ABD ≅ Δ BAC
(ii) BD = AC
(iii) ∠ ABD = ∠ BAC
Proof: (i) In Δ ABD and Δ BAC
AD = BC [given]
∠ DAB = ∠ CBA [given]
AB = AB [Common]
By SAS Congruency Criterion Rule
Δ ABD ≅ Δ BAC
(ii) BD = AC [CPCT]
(iii) ∠ ABD = ∠ BAC [CPCT]
Q3. AD and BC are equal perpendiculars to a line segment AB (see the given figure). Show that CD bisects AB.
Solution:
Given: AD and BC are equal perpendiculars to a line segment AB.
To prove: CD bisects AB.
Proof:
In ∆BOC and ∆AOD
∠ BOC = ∠AOD (Vertically opposite angles)
∠CBO = ∠DAO (Each 90º)
BC = AD (Given)
By AAS Congruence Criterion Rule
∆BOC ≅ ∆AOD
BO = AO (By CPCT)
Hence, CD bisects AB.
Q4. l and m are two parallel lines intersected by another pair of parallel lines p and q (See the given figure). Show that ∆ABC ≅ ∆CDA
Solution:
Given: l and m are two parallel lines intersected by another pair of parallel lines p and q.
To prove: ∆ABC ≅ ∆CDA
Proof:
In ∆ABC and ∆CDA,
∠ BAC = ∠DCA (Alternate interior angles, as p || q)
AC = CA (Common)
∠ BCA = ∠DAC (Alternate interior angles, as l || m)
By AAS Congruence Criterion Rule
∆ABC ≅ ∆CDA
Q5. 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 the given figure). Show that: (i) ∆APB ≅ ∆AQB (ii) BP = BQ or B is equidistant from the arms of ∠A.
Solution:
Given: 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.
To prove:
(i) ∆APB ≅ ∆AQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.
Proof:
In ∆APB and ∆AQB,
∠ APB = ∠AQB (Each 90º)
∠ PAB = ∠QAB (l is the angle bisector of A)
AB = AB (Common)
By AAS Congruence Criterion Rule
∆APB ≅ ∆AQB
BP = BQ [CPCT]
it can be said that B is equidistant from the A.
Q6. In the given figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
Solution:
Given: AC = AE, AB = AD and ∠BAD = ∠EAC.
To prove: BC = DE.
Proof: ∠BAD = ∠EAC
BAD + DAC = EAC + DAC
BAC = DAE
In ∆BAC and ∆DAE
AC = AE (Given)
AB = AD (Given)
∠BAC = ∠DAE (proved above)
By SAS Congruence Criterion Rule
∆BAC ≅ ∆DAE
BC = DE (CPCT)
Q7. 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 the given figure).
Show that:
(i) ∆DAP ≅ ∆EBP
(ii) AD = BE
Solution:
Given: 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.
To prove:
(i) ∆DAP ≅ ∆EBP
(ii) AD = BE
Proof: In ∆ DPA and ∆ EPB
∠EPA = ∠DPB
EPA + DPE = DPB + DPE
∠ DPA = ∠EPB
∠BAD =∠ ABE (Given)
∠EPA = ∠DPB (Given)
AP =BP (P is the midpoint of AB)
By AAS Congruence Criterion Rule
∆DAP ≅ ∆EBP
AD = BE (CPCT)
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