NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.3

NCERT Solutions for Class 9 Maths Chapter 7 Triangles Exercise 7.3 are given here in simple, step-by-step form. NCERT stands for National Council of Educational Research and Training, the body that makes your textbook. Chapter 7 teaches congruence, which means two triangles are exactly equal in shape and size. Exercise 7.1 covered the SAS and ASA rules, and Exercise 7.2 covered the properties of an isosceles triangle. Exercise 7.3 brings in one more rule that works only for right-angled triangles, called the RHS rule.

This exercise has 5 questions. All of them ask you to prove something. You will work with two isosceles triangles drawn on the same base, with an altitude of an isosceles triangle, with medians of two triangles, and with two equal altitudes. Students can use this page for homework and revision, parents can use it to check the work at home, and teachers can use it in class. A free printable PDF of these solutions is also available for offline study.

Exercise 7.3 – Questions and Answers

Q 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 the figure below). 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.

Answer:
Given: ΔABC and ΔDBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC.
To Prove: (i) ΔABD ≅ ΔACD (ii) ΔABP ≅ ΔACP (iii) AP bisects ∠A as well as ∠D. (iv) AP is the perpendicular bisector of BC.

Proof:
(i) In ΔABD and ΔACD
AB = AC [Given]
DB = DC [Given]
AD = AD [Common]
∴ ΔABD ≅ ΔACD [By S.S.S.]
∠BAD = ∠CAD [By C.P.C.T.]

(ii) In ΔABP and ΔACP
AB = AC [Given]
∠BAP = ∠CAP [Proved above]
AP = AP [Common]
∴ ΔABP ≅ ΔACP [By S.A.S.]
∠BAP = ∠CAP [By C.P.C.T.]
BP = CP [By C.P.C.T.]

(iii) ∵ ∠BAP = ∠CAP [By C.P.C.T.]
⇒ AP bisects ∠A.

In ΔBDP and ΔCDP
BD = CD [Given]
DP = DP [Common]
BP = CP [By C.P.C.T.]
∴ ΔBDP ≅ ΔCDP [By S.S.S.]
∠BDP = ∠CDP [By C.P.C.T.]
⇒ DP bisects ∠BDC, i.e., ∠D.

Thus, AP bisects ∠A as well as ∠D.

(iv) ∠APB = ∠APC [By C.P.C.T.]
and ∠APB + ∠APC = 180° [Linear pair]
⇒ ∠APB + ∠APB = 180°
⇒ 2∠APB = 180°
⇒ ∠APB = 180°/2 = 90°
And BP = CP

So, AP is the perpendicular bisector of BC.

Q 2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that:
(i) AD bisects BC
(ii) AD bisects ∠A.

Answer:
Given: AD is the altitude in isosceles ΔABC in which AB = AC.
To Prove: (i) AD bisects BC (ii) AD bisects ∠A

Proof:
(i) In ΔABD and ΔACD
AB = AC [Given]
∠ADB = ∠ADC [Each 90°]
AD = AD [Common]
ΔABD ≅ ΔACD [By R.H.S.]
So, BD = CD [By C.P.C.T.]
Therefore, AD bisects BC.

(ii) ∠BAD = ∠CAD [By C.P.C.T.]
Therefore, AD bisects ∠A.

Hence proved.

Q 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 below). Show that:
(i) ΔABM ≅ ΔPQN
(ii) ΔABC ≅ ΔPQR

Answer:
Given: In ΔABC and ΔPQR, AB = PQ, BC = QR and AM = PN.
To Prove: (i) ΔABM ≅ ΔPQN (ii) ΔABC ≅ ΔPQR

Proof:
(i) In ΔABM and ΔPQN
AB = PQ [Given]
BM = QN [½BC = ½QR]
AM = PN [Given]
∴ ΔABM ≅ ΔPQN [By S.S.S.]
∠ABM = ∠PQN [By C.P.C.T.]
⇒ ∠B = ∠Q

(ii) In ΔABC and ΔPQR
AB = PQ [Given]
∠B = ∠Q [Proved above]
BC = QR [Given]
∴ ΔABC ≅ ΔPQR [By S.A.S.]

Hence proved.

Q 4. BE and CF are two equal altitudes of a triangle ABC. Show that:
(i) ΔABE ≅ ΔACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.

Answer:

Given: In ΔABC, BE ⊥ AC and CF ⊥ AB and BE = CF.
To Prove: ΔABC is an isosceles triangle.

Proof:
(i) In ΔBCE and ΔCBF
∠BEC = ∠CFB [Each 90°]
BC = BC [Common]
BE = CF [Given]
∴ ΔBCE ≅ ΔCBF [By R.H.S.]

(ii) AB = AC [By C.P.C.T.]
i.e., ABC is an isosceles triangle.

Q 5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.

Answer:
Given: In ΔABC, AB = AC and AP ⊥ BC.
To Prove: ΔABC is an isosceles triangle.

Proof: In ΔABP and ΔACP
∠APB = ∠APC [Each 90°]
AB = AC [Given]
AP = AP [Common]
ΔABP ≅ ΔACP [By R.H.S.]
∴ ∠ABP = ∠ACP [By C.P.C.T.]
⇒ ∠B = ∠C

Hence proved.

Related Links

• NCERT Solutions Class 9 Maths Chapter 7 Exercise 7.1
• NCERT Solutions Class 9 Maths Chapter 7 Exercise 7.2
• NCERT Solutions Class 9 Maths Chapter 7 – Triangles