NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3
NCERT Solutions for Class
9 Maths Chapter 7 Triangles Ex 7.3 are the part of NCERT Solutions for Class 9 Maths.
Here you can find the NCERT Solutions for Chapter 7 Triangles Ex 7.3.
- NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1
- NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.2
- NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3
- NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.4
- NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.5
Ex 7.3 Class 9 Maths Question 1.
ΔABC and ΔDBC 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)
ΔABD ≅ ΔACD
(ii)
ΔABP ≅ ΔACP
(iii)
AP bisects ∠A as well as ∠D
(iv) AP
is the perpendicular bisector of BC.
Solution:
(i) In ∆ABD and ∆ACD, we have
AB = AC [Given]
AD = DA [Common]
BD = CD [Given]
∴ ∆ABD ≅ ∆ACD [By SSS
congruency rule]
∠BAD = ∠CAD [By CPCT] …(1)
(ii)
In ∆ABP and ∆ACP, we have
AB = AC [Given]
∠BAP = ∠CAP [From (1)]
∴ AP = PA [Common]
∴ ∆ABP ≅ ∆ACP [By SAS
congruency rule]
(iii)
Since, ∆ABP ≅ ∆ACP
⇒ ∠BAP = ∠CAP [By CPCT]
∴ AP is the bisector of ∠A.
Again, in ∆BDP and ∆CDP, we have
BD = CD [Given]
DP = PD [Common]
BP = CP [ ∵ ∆ABP ≅ ∆ACP]
⇒ ∆BDP = ∆CDP [By SSS congruency rule]
∴ ∠BDP = ∠CDP [By CPCT]
⇒ DP (or AP) is the bisector of ∠BDC.
∴ AP is the bisector of ∠A as well as ∠D.
(iv)
Since ∆ABP ≅ ∆ACP
⇒ ∠APB = ∠APC and BP = CP [By
CPCT]
But ∠APB + ∠APC = 180° [Linear
Pair]
∴ ∠APB = ∠APC = 90°
⇒ AP ⊥ BC, also BP = CP
Hence, AP is the perpendicular bisector of BC.
Ex 7.3 Class 9 Maths 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:
(i) In right-angled ∆ABD and ∆ACD, we have
AB = AC [Given]
∠ADB = ∠ADC [Each 90°]
AD = DA [Common]
∴ ∆ABD ≅ ∆ACD [By RHS congruency
rule]
So, BD = CD [By CPCT]
⇒ D is the mid-point of BC or AD bisects BC.
(ii)
Since ∆ABD ≅ ∆ACD
⇒ ∠BAD = ∠CAD [By CPCT]
So, AD bisects ∠A.
Ex 7.3 Class 9 Maths 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:
In ∆ABC, AM is the median.
∴ BM = ½ BC ……(1)
In ∆PQR, PN is the median.
∴ QN = ½ QR ……(2)
And BC = QR [Given]
⇒ ½ BC = ½ QR
⇒ BM = QN ……(3) [From eqs. (1) and (2)]
(i)
In ∆ABM and ∆PQN, we have
AB = PQ [Given]
AM = PN [Given]
BM = QN [From eq. (3)]
∴ ∆ABM ≅ ∆PQN [By SSS
congruency rule]
(ii)
Since ∆ABM ≅ ∆PQN
⇒ ∠B = ∠Q ……(4) [By CPCT]
Now, in ∆ABC and ∆PQR, we have
∠B = ∠Q [From
eq. (4)]
AB = PQ [Given]
BC = QR [Given]
∴ ∆ABC ≅ ∆PQR [By SAS congruency
rule]
Ex 7.3 Class 9 Maths 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:
Since BE ⊥ AC [Given]
∴ BEC is a right-angled triangle such that ∠BEC = 90°
Similarly, ∠CFB = 90°
Now, in right-angled ∆BEC and ∆CFB, we have
BE = CF [Given]
BC = CB [Common
hypotenuse]
∠BEC = ∠CFB [Each 90°]
∴ ∆BEC ≅ ∆CFB [By RHS
congruency rule]
So, ∠BCE = ∠CBF [By CPCT]
or ∠BCA = ∠CBA
Now, in ∆ABC, ∠BCA = ∠CBA
⇒ AB = AC [Sides
opposite to equal angles of a ∆ are equal]
∴ ∆ABC is an isosceles triangle.
Ex 7.3 Class 9 Maths Question 5.
ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.
Solution:
We have, AP ⊥ BC [Given]
∠APB = 90° and ∠APC = 90°
In ∆ABP and ∆ACP, we have
∠APB = ∠APC [Each 90°]
AB = AC [Given]
AP = AP [Common]
∴ ∆ABP ≅ ∆ACP [By RHS congruency rule]
So, ∠B = ∠C [By CPCT]
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