7. Triangles Mathematics Exercise - 7.3 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 7.3 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
3. Exercise 7.3
EXERCISE- 7.3
1.Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that:
(i) Δ ABD ≅ Δ ACD
(ii) Δ ABP ≅ Δ ACP
(iii) AP bisects ∠ A as well as ∠ D.
(iv) AP is the perpendicular bisector of BC.
Solution:
Given: Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC.
To prove:
(i) Δ ABD ≅ Δ ACD
(ii) Δ ABP ≅ Δ ACP
(iii) AP bisects ∠ A as well as ∠ D.
(iv) AP is the perpendicular bisector of BC
Proof:
In ΔABD and Δ ACD
AB = AC [given]
BD = CD [given]
AD = AD [common]
By SSS Congruence Criterion Rule
Δ ABD ≅ Δ ACD
∠ BAD = ∠CAD [CPCT]
∠ BAP = ∠CAP [CPCT] …
(ii)In ΔABP and Δ ACP
AB = AC [given]
∠ BAP = ∠CAP [proved above]
AP = AP [common]
By SAS Congruence Criterion Rule
Δ ABP ≅ Δ ACP
BP = CP [CPCT] … 2
∠APB = ∠APC [CPCT]
(iii) ∠ BAP = ∠CAP [From eq. 1]
Hence, AP bisects ∠ A.
Now, In Δ BDP and Δ CDP
BD = CD [given]
BP = CP [given]
DP = DP [common]
By SSS Congruence Criterion Rule
Δ BDP ≅ Δ CDP
∠ BDP = ∠CDP [CPCT]
AP bisects ∠ D.
(iv) AP stands on B
∠APB + ∠APC = 1800
∠APB +∠APB = 1800[proved above]
∠APB = 1800 /2
∠APB = 900
AP is the perpendicular bisector of BC.
2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC (ii) AD bisects ∠ A.
Solution:
Given: AD is an altitude of an isosceles triangle ABC in which AB = AC.
To prove: (i) AD bisects BC
(ii) AD bisects ∠ A.
Proof: In ∆BAD and ∆CAD
∠ ADB = ∠ADC (Each 90º as AD is an altitude)
AB = AC (Given)
AD = AD (Common)
By RHS Congruence Criterion Rule
∆BAD ≅ ∆CAD
BD = CD (By CPCT)
Hence, AD bisects BC.
∠BAD = ∠CAD (By CPCT)
Hence, AD bisects ∠ A
3. Two sides AB and BC and median AM of one triangle ABC are respectively
equal to sides PQ and QR and median PN of Δ PQR (see Fig. 7.40). Show that:
(i) Δ ABM ≅ Δ PQN
(ii) Δ ABC ≅ Δ PQR
Solution:
Given: Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of Δ PQR.
To prove: (i) Δ ABM ≅ Δ PQN
(ii) Δ ABC ≅ Δ PQR
Proof: In ∆ABC, AM is the median to BC.
BM = 1/2 BC ... 1
In ∆PQR, PN is the median to QR.
QN = 1/2 QR ... 2
from eq .1 & 2
BM = QN ... 3
Now in ABM and PQN
AB = PQ (Given)
BM = QN [From equation (3)]
AM = PN [given]
By SSS congruence Criterion rule
∆ABM ≅ ∆PQN
∠B =∠Q [CPCT]
Now in∆ ABC and∆ PQR
AB = PQ [given]
∠B = ∠Q [prove above ]
BC = QR [given]
By SAS congruence Criterion rule
∆ ABC ≅ ∆ PQR
4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Solution:
Given: BE and CF are two equal altitudes of a triangle ABC.
To prove: ABC is a isosceles.
Proof: In ∆BEC and ∆CFB,
BE = CF (Given)
∠BEC = CFB (Each 90°)
BC = CB (Common)
By RHS congruence Criterion rule
∆BEC ≅ ∆CFB
∠BCE = ∠CBF (By CPCT)
AB = AC [Sides opposite to equal angles of a triangle are equal]
Hence, ABC is isosceles.
5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that:
∠ B = ∠ C.
Solution:
Given: ABC is an isosceles triangle with AB = AC.
To prove: ∠ B = ∠ C.
Construction: Draw AP ⊥ BC to
Proof : In ∆APB and ∆APC
∠APB = ∠APC (Each 90º)
AB =AC (Given)
AP = AP (Common)
By RHS Congruence Criterion Rule
∆APB ≅ ∆APC
∠B = ∠C [CPCT]
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