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# Ex 7.3 NCERT Solutions- Triangles Class 9 Notes | EduRev

## Mathematics (Maths) Class 9

Created by: Full Circle

## Class 9 : Ex 7.3 NCERT Solutions- Triangles Class 9 Notes | EduRev

The document Ex 7.3 NCERT Solutions- Triangles Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9

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 ΔABD and ΔACD, we have
AB = AC                   [Given]
BD = CD                  [Given]
∴ ΔABD ≌ ΔACD                   [SSS Criteria]

(ii) In ΔABP and ΔACP, we have
AB = AC                   [Given]
∴ AB = AC
⇒ ∠B = ∠C        [∵ Angle opposite to equal sides are equal]
AP = AP                  [Common]
∴ ΔABP ≌ ΔACP                   [SAS Criteria]

(iii) Since, ΔABP ≌ ΔACP
∴ Their corresponding parts are congruent.
⇒ ∠BAP = ∠CAP
∴ AP is the bisector of ∠A                  . ...(1)
Again, in ΔBDP and ΔCDP, we have
BD = CD                   [Given]
∠DBP = ∠CDP         [Angles opposite to equal sides]
DP = DP                  [Common]
⇒ ∠BDP ≌ ∠CDP
∴ ∠BDP = ∠CDP                  [c.p.c.t.]
⇒ DP (or AP) is the bisector of ∠D                  . ...(2)
From (1) and (2), AP is the bisector of ∠A as well as ∠D.

(iv) ∵ ΔABP ≌ ΔACP
∴ Their corresponding parts are equal.
⇒ ∠APB = ∠APC
But ∠APB + ∠APC = 180º                          [Linear pair]
∴ ∠APB = ∠APC = 90º
⇒  AP ⊥ BC
⇒ AP is the perpendicular bisector of BC

Question 2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
Solution:
(i) In ΔABD and ΔACD, we have
AB = AC                   [Given]
∠B= ∠C
[Angles opposite to equal sides] ∴ ΔABD ≌ ΔACD
⇒ Their corresponding parts are equal.
∴ BD = CD
⇒ D is the mid-point of BC or AD bisects BC.

(ii) Since, ΔABD ≌ ΔACD,
∴ Their corresponding parts are congruent.

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 ΔPQR (see figure). Show that (i) ΔABM ≌ ΔPQN
(ii) ΔABC ≌ ΔPQR Solution: In ΔABC, AM is a median                   [Given].
∴ BM = (1/2)BC                   …(1)
In ΔPQR, PN is a median.
∴ QN = (1/2)QR                   …(2)
∵ BC = QR [Given]
∴ (1/2)BC = (1/2)QR
⇒ BM = QN
[From (1) and (2)]

(i) In ΔABM and ΔPQN, we have
∵ AB = PQ                   [Given]
AM = PN                   [Given]
BM = QN                  [Proved]
∴ ΔABM ≌ ΔPQN                  [SSS criteria]

(ii) ∵ ΔABM ≌ ΔPQN
∴ Their corresponding parts are congruent.
⇒ ∠B= ∠Q
Now, in ΔABC and ΔPQR, we have
∠B= ∠Q                  [Proved]
AB = PQ                   [Given]
BC = QR                  [Given]
∴ ΔABC ≌ ΔPQR

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:
∵ BE ⊥ AC                  [Given]
∴ ΔBEC is a right triangle such that ∠BEC = 90º
Similarly, ∠CFB = 90º
Now, in right ΔBEC and right ΔCFB, we have BE = CF                  [Given]
BC = CB                  [common]
∴ Using RHS criteria,
ΔBEC ≌ ΔCFB
∴ Their corresponding parts are equal.
⇒ ∠BCE = ∠CBF or ∠BCA = ∠CBA
Now, in ΔABC, ∠BCA = ∠CBA
∴ Their opposite sides are equal.
⇒ AB = AC
∴ ΔABC 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]
∴ ∠APB = 90º and APC = 90º
In ΔABP and ΔACP, we have ∠APB = ∠APC               [each = 90º]
AB = AC                         [Given]
AP = AP                         [common]
∴ Using RHS criteria,
ΔABP ≌ ΔACP
∴ Their corresponding parts are congruent.
⇒ ∠B= ∠C

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