class 9 maths exercise 7.3 questions no. 3
Answers
Step-by-step explanation:
(i) in triangle ABM and triangle PQN
side AB. congruent to SIDE PQ
side BM congruent to SIDE QN
Median AM Cong to median PN
triangle ABM congruent triangle PQN
( SSS TEST)
NOTE : I'm not sure about it....u May refer to other answers too...
Exercise 7.3
Question 1 | Page 128
∆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 the given 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 ∆ABD and ∆ACD,
AB = AC (Given)
BD = CD (Given)
AD = AD (Common)
∆ABD ≅ ∆ACD (By SSS congruence rule)
∠BAD = ∠CAD (By CPCT)
∠BAP = ∠CAP …. (1)
(ii) In ∆ABP and ∆ACP,
AB = AC (Given)
∠BAP = ∠CAP [From equation (1)]
AP = AP (Common)
∴ ∆ABP ≅ ∆ACP (By SAS congruence rule)
∴ BP = CP (By CPCT) … (2)
(iii) From Equation (1),
∠BAP = ∠CAP
Hence, AP bisects ∠A.
In ∆BDP and ∆CDP,
BD = CD (Given)
DP = DP (Common)
BP = CP [From equation (2)]
∴ ∆BDP ≅ ∆CDP (By SSS Congruence rule)
∴ ∠BDP = ∠CDP (By CPCT) … (3)
Hence, AP bisects ∠D.
(iv) ∆BDP ≅ ∆CDP
∴ ∠BPD = ∠CPD (By CPCT) …. (4)
∠BPD + ∠CPD = 180o (Linear pair angles)
∠BPD + ∠BPD = 180o
2∠BPD = 180o [From Equation (4)]
∠BPD = 90o … (5)
From Equations (2) and (5), it can be said that AP is the perpendicular bisector of BC.
Question 2 | Page 128
AD is an altitude of an isosceles triangles ABC in which AB = AC. Show that
(i) AD bisects BC (ii) AD bisects ∠A.
Solution:
(i) In ∆BAD and ∆CAD,
∠ADB = ∠ADC (Each 90º as AD is an altitude)
AB = AC (Given)
AD = AD (Common)
∴ ∆BAD ≅ ∆CAD (By RHS Congruence rule)
∴ BD = CD (By CPCT)
Hence, AD bisects BC.
(ii) Also, by CPCT,
∠BAD = ∠CAD
Hence, AD bisects ∠A.
Question 3 | Page 128
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 the given figure). Show that:
(i) ∆ABM ≅ ∆PQN
(ii) ∆ABC ≅ ∆PQR
Solution:
(i) In ∆ABC, AM is the median to BC.
∴ BM = BC
In ∆PQR, PN is the median to QR.
∴ QN = QR
However, BC = QR
∴ BC = QR
∴ BM = QN … (1)
In ∆ABM and ∆PQN,
AB = PQ (Given)
BM = QN [From Equation (1)]
AM = PN (Given)
∆ABM ≅ ∆PQN (By SSS congruence rule)
∠ABM = ∠PQN (By CPCT)
∠ABC = ∠PQR … (2)
(ii) In ∆ABC and ∆PQR,
AB = PQ (Given)
∠ABC = ∠PQR [From Equation (2)]
BC = QR (Given)
∴ ∆ABC ≅ ∆PQR (By SAS congruence rule)
Question 4 | Page 128
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Solution:
In ∆BEC and ∆CFB,
∠BEC = ∠CFB (Each 90°)
BC = CB (Common)
BE = CF (Given)
∴ ∆BEC ≅ ∆CFB (By RHS congruency)
∴ ∠BCE = ∠CBF (By CPCT)
∴ AB = AC (Sides opposite to equal angles of a triangle are equal)
Hence, ∆ABC is isosceles.
Question 5 | Page 128
ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.
Solution:
In ∆APB and ∆APC,
∠APB = ∠APC (Each 90º)
AB =AC (Given)
AP = AP (Common)
∴ ∆APB ≅ ∆APC (Using RHS congruence rule)
∴ ∠B = ∠C (By using CPCT)
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