CLASS 9 MATH NCERT SOLUTION FOR CHAPTER – 7 TRIANGLES EX – 7.3

Triangles

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
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3 Q1

(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, we have
AB = AC [Given]
AD = DA [Common]
BD = CD [Given]
∴ ∆ABD ≅ ∆ACD [By SSS congruency]
∠BAD = ∠CAD [By C.P.C.T.] …(1)

(ii) In ∆ABP and ∆ACP, we have
AB = AC [Given]
∠BAP = ∠CAP [From (1)]
∴ AP = PA [Common]
∴ ∆ABP ≅ ∆ACP [By SAS congruency]

(iii) Since, ∆ABP ≅ ∆ACP
⇒ ∠BAP = ∠CAP [By C.P.C.T.]
∴ AP is the bisector of ∠A.
Again, in ∆BDP and ∆CDP,
we have BD = CD [Given]
DP = PD [Common]
BP = CP [ ∵ ∆ABP ≅ ∆ACP]
⇒ A BDP = ACDP [By SSS congruency]
∴ ∠BDP = ∠CDP [By C.P.C.T.]
⇒ DP (or AP) is the bisector of ∠BDC
∴ AP is the bisector of ∠A as well as ∠D.

(iv) As, ∆ABP ≅ ∆ACP
⇒ ∠APS = ∠APC, BP = CP [By C.P.C.T.]
But ∠APB + ∠APC = 180° [Linear Pair]
∴ ∠APB = ∠APC = 90°
⇒ AP ⊥ BC, also BP = CP
Hence, AP is the 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:
(i) In right ∆ABD and ∆ACD, we have
AB =AC [Given]
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3 Q2
∠ADB = ∠ADC [Each 90°]
AD = DA [Common]
∴ ∆ABD ≅ ∆ACD [By RHS congruency]
So, BD = CD [By C.P.C.T.]
⇒ D is the mid-point of BC or AD bisects BC.

(ii) Since, ∆ABD ≅ ∆ACD,
⇒ ∠BAD = ∠CAD [By C.P.C.T.]
So, AD bisects ∠A

Question 3.
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and OR and median PN of ∆PQR (see figure). Show that

(i) ∆ABC ≅ ∆PQR
(ii) ∆ABM ≅ ∆PQN

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3 Q3

Solution:
In ∆ABC, AM is the median.
∴BM = 12 BC ……(1)
In ∆PQR, PN is the median.
∴ QN = 12QR …(2)
And BC = QR [Given]
⇒ 12BC = 12QR
⇒ BM = QN …(3) [From (1) and (2)]

(i) In ∆ABM and ∆PQN, we have
AB = PQ , [Given]
AM = PN [Given]
BM = QN [From (3)]
∴ ∆ABM ≅ ∆PQN [By SSS congruency]

(ii) Since ∆ABM ≅ ∆PQN
⇒ ∠B = ∠Q …(4) [By C.P.C.T.]
Now, in ∆ABC and ∆PQR, we have
∠B = ∠Q [From (4)]
AB = PQ [Given]
BC = QR [Given]
∴ ∆ABC ≅ ∆PQR [By SAS congruency]

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]
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3 Q4
∴ BEC is a right triangle such that ∠BEC = 90°
Similarly, ∠CFB = 90°
Now, in right ∆BEC and ∆CFB, we have
BE = CF [Given]
BC = CB [Common hypotenuse]
∠BEC = ∠CFB [Each 90°]
∴ ∆BEC ≅ ∆CFB [By RHS congruency]
So, ∠BCE = ∠CBF [By C.P.C.T.]
or ∠BCA = ∠CBA
Now, in ∆ABC, ∠BCA = ∠CBA
⇒ AB = AC [Sides opposite to equal angles of a ∆ 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:
We have, AP ⊥ BC [Given]
NCERT Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.3 Q5
∠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]
So, ∠B = ∠C [By C.P.C.T.]

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