BSEB Class 9 Maths Chapter 7 Triangles Ex 7.3 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 7 Triangles Ex 7.3 Book Answers ~ HSSlive: Plus One & Plus Two Notes & Solutions for Kerala State Board

BSEB Class 9 Maths Chapter 7 Triangles Ex 7.3 Textbook Solutions PDF: Download Bihar Board STD 9th Maths Chapter 7 Triangles Ex 7.3 Book Answers

BSEB Class 9th Maths Chapter 7 Triangles Ex 7.3 Textbooks Solutions and answers for students are now available in pdf format. Bihar Board Class 9th Maths Chapter 7 Triangles Ex 7.3 Book answers and solutions are one of the most important study materials for any student. The Bihar Board Class 9th Maths Chapter 7 Triangles Ex 7.3 books are published by the Bihar Board Publishers. These Bihar Board Class 9th Maths Chapter 7 Triangles Ex 7.3 textbooks are prepared by a group of expert faculty members. Students can download these BSEB STD 9th Maths Chapter 7 Triangles Ex 7.3 book solutions pdf online from this page.

Bihar Board Class 9th Maths Chapter 7 Triangles Ex 7.3 Textbooks Solutions PDF

Bihar Board STD 9th Maths Chapter 7 Triangles Ex 7.3 Books Solutions with Answers are prepared and published by the Bihar Board Publishers. It is an autonomous organization to advise and assist qualitative improvements in school education. If you are in search of BSEB Class 9th Maths Chapter 7 Triangles Ex 7.3 Books Answers Solutions, then you are in the right place. Here is a complete hub of Bihar Board Class 9th Maths Chapter 7 Triangles Ex 7.3 solutions that are available here for free PDF downloads to help students for their adequate preparation. You can find all the subjects of Bihar Board STD 9th Maths Chapter 7 Triangles Ex 7.3 Textbooks. These Bihar Board Class 9th Maths Chapter 7 Triangles Ex 7.3 Textbooks Solutions English PDF will be helpful for effective education, and a maximum number of questions in exams are chosen from Bihar Board.

Bihar Board Class 9th Maths Chapter 7 Triangles Ex 7.3 Books Solutions

Board	BSEB
Materials	Textbook Solutions/Guide
Format	DOC/PDF
Class	9th
Subject	Maths Chapter 7 Triangles Ex 7.3
Chapters	All
Provider	Hsslive

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BSEB Bihar Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.3

Question 1.
∆ ABC and ∆ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see figure). If AD is extended to intersect BC at P, show that

(i) ∆ ABD ≅ ∆ ACD
(ii) ∆ ABP ≅ ∆ ACP
(iii) AP bisects ∠A as well as ∠D.
(iv) AP is the perpendicular bisector of BC.
Solution:
(i) In ∆s ABD and DBC, we have
AB = AC [Given]
BD = DC [Given]
and, AD = AD [Common]
∴ By SSS criterion of congruence, we have
∆ ABD ≅ ∆ ACD.

(ii) In ∆s ABP and ACP, we have
AB = AC [Given]
∠BAP = ∠PAC
[∵ ∆ ABD = ∆ ACD ⇒ ∠B AD = ∠DAC ⇒ ∠BAP = ∠PAC]
and, AP = AP [Common]
∴ By SAS criterion 6f congruence, we have
∆ ABP ≅ ∆ ACP.

(iii) Since ∆ ABD = ∆ ACD. Therefore,
∠BAD = ∠DAC
⇒ AD bisects ∠A ⇒ AP bisects ∠A … (1)
In ∆s BDP and CDP, we have
BD = CD [Given]
BP = PC [∵ ∆ ASP = ∆ ACP ⇒ BP = PC]
and DP = DP [Common]
∴ By SSS criterion of congruence, we have
∆ BDP ≅ ∆ CDP
⇒ ∠BDP = ∠PDC
⇒ DP bisects ∠D ⇒ AP bisects ∠D … (2)
Combining (1) and (2), we get
AP bisects ∠A as well as ∠D.

(iv) Since AP stands on BC
∴ ∠APB + ∠APC = 180° [Linear Pair]
But ∠APB = ∠APC [Proved above]
∴ ∠APB = ∠APC = 180°2 = 90°
Also , BP = PC [Proved above]
⇒ AP is perpendicular bisector of BC.

Question 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:
AD is the altitude drawn from vertex A of an isosceles ∆ ABC to the opposite base BC so that AB = AC, ∠ADC = ∠ADB = 90°.
Now, in As ADB and ADC, we have

Hyp. AB = Hyp. AC [Given]
AD = AD [Common]
and ∠ADC = ∠ADB [∵ Each = 90°]
∴ By RHS criterion of congruence, we have
∆ ADB ≅ ∆ ADC
⇒ BD = DC and ∠BAC = ∠DAC
[∵ Corresponding parts of congruent triangles are equal]
Hence, AD bisects BC, which proves (i), and AD bisects ∠A, which proves (ii).

Question 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 ∆ PQR (see figure). Show that:
(i) ∆ ABM ≅ ∆ PQN
(ii) ∆ ABC ≅ ∆ PQR.

Solution:
Two ∆s ABC and PQR in which AB = PQ, BC = QR and AM = PN.
Since AM and PN are medians of As ABC and PQR respectively.
Now, BC = QR [Given]
⇒ 12BC = 12QR ⇒ BM = QN … (1)
Now, in ∆s ABM and PQN, we have
AB = PQ [Given]
BM = QN [From (i)]
and, AM = PN [Given]
∴ By SSS criterion of congruence, we have ∆ ABM = ∆ PQN, which proves (i)
⇒ ∠B = ∠Q … (2)
[∵ Corresponding parts of congruent triangles are equal]
Now, in ∆s ABC and PQR, we have
AB = PQ [Given]
∠B = ∠Q [From (2)]
BC = QR [Given]
∴ By SAS criterion of congruence, we have
∆ ABC = ∆ PQR, which proves (ii).

Question 4.
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Solution:
In ∆s BCF and CBE, we have

∠BFC = ∠CEB [∵ Each = 90°]
Hyp. BC = Hyp. BC [Common]
FC = EB
∴ By R.H.S. criterion of congruence, we have
∆ BCF ≅ ∆ CBE
⇒ ∠FBC = ∠ECB
[∵ Corresponding parts of congruent triangles are equal]
Now, in ∆ ABC
∠ABC = ∠ACB [∵∠FBC = ∠ECB]
⇒ AB = AC [∵ Sides opposite to equal angles of a triangle are equal]
∴ ∆ ABC is an isosceles triangle.

Question 5.
ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.
Solution:
In ∆s ABP and ACP, we have

AB = AC [Given]
AP = AP [Common]
and ∠APB = ∠APC [∵ Each = 90°]
∴ By R.H.S. criterion of congruence, we have
∆ ABP ≅ ∆ ACP
⇒ ∠B = ∠C [∵ Corresponding parts of congruent triangles are equal]

BSEB Textbook Solutions PDF for Class 9th

Bihar Board Class 9th Maths Chapter 7 Triangles Ex 7.3 Textbooks for Exam Preparations

Bihar Board Class 9th Maths Chapter 7 Triangles Ex 7.3 Textbook Solutions can be of great help in your Bihar Board Class 9th Maths Chapter 7 Triangles Ex 7.3 exam preparation. The BSEB STD 9th Maths Chapter 7 Triangles Ex 7.3 Textbooks study material, used with the English medium textbooks, can help you complete the entire Class 9th Maths Chapter 7 Triangles Ex 7.3 Books State Board syllabus with maximum efficiency.

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