Looking for accurate SEBA Class 9 Maths Exercise 7.3 (Triangles) textual solutions for the 2026–27 academic year? Access complete step-by-step geometry proofs covering SSS (Side-Side-Side) and RHS (Right angle-Hypotenuse-Side) triangle congruence criteria to score top marks in your annual exams.
EXERCISE 7.3
1. Δ ABC and Δ DBC are two isosceles Δ on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.41). If AD is extended to intersect BC at P, show that:
(i) Δ ABD ≅ Δ ACD
(i) To prove: Δ ABD ≅ Δ ACD
Proof:
In Δ ABD and Δ ACD,
AB = AC (Given, Δ ABC is isosceles)
BD = CD (Given, Δ DBC is isosceles)
AD = AD (Common side)
By SSS rule,
Δ ABD ≅ Δ ACD
Therefore,
∠BAD = ∠CAD — (1) [CPCT]
∠BDA = ∠CDA — (2) [CPCT]
(ii) Δ ABP ≅ Δ ACP
To prove: Δ ABP ≅ Δ ACP
Proof:
In Δ ABP and Δ ACP,
AB = AC (Given)
∠BAP = ∠CAP [From (1)]
AP = AP (Common side)
By SAS rule,
Δ ABP ≅ Δ ACP
Therefore,
BP = CP — (3) [CPCT]
∠APB = ∠APC — (4) [CPCT]
(iii) AP bisects ∠A as well as ∠D.
To prove: AP bisects ∠A as well as ∠D
Proof:
From eq (1), ∠BAP = ∠CAP.
Thus, AP bisects ∠A.
Now,
∠BDA + ∠BDP = 180° (Linear pair)
∠BDP = 180° – ∠BDA
Also,
∠CDA + ∠CDP = 180° (Linear pair)
∠CDP = 180° – ∠CDA
Since ∠BDA = ∠CDA [From (2)],
∠BDP = ∠CDP
Thus, AP bisects ∠D.
(iv) AP is the perpendicular bisector of BC.
To prove: AP is the perpendicular bisector of BC
Proof:
From eq (3),
BP = CP
From eq (4),
∠APB = ∠APC
Since BPC is a straight line,
∠APB + ∠APC = 180° (Linear pair)
∠APB + ∠APB = 180°
2∠APB = 180°
∠APB = 90°
Therefore ∠APB = ∠APC = 90°
Hence, AP is the perpendicular bisector of BC.
2. AD is an altitude of an isosceles Δ ABC in which AB = AC. Show that:
(i) AD bisects BC
To prove: AD bisects BC (i.e., BD = CD)
Proof:
In right Δs Δ ADB and Δ ADC,
∠ADB = ∠ADC = 90° (AD is altitude)
AB = AC (Hypotenuse, Given)
AD = AD (Side, Common)
By RHS rule,
Δ ADB ≅ Δ ADC
Therefore BD = CD [CPCT]
Hence, AD bisects BC.
(ii) AD bisects ∠A.
To prove: AD bisects ∠A
Proof:
Since Δ ADB ≅ Δ ADC,
∠BAD = ∠CAD [CPCT]
Hence, AD bisects ∠A.
SEBA Class 9 Maths Exercise 7.3 Solutions
3. Two sides AB and BC and median AM of one Δ ABC are respectively equal to sides PQ and QR and median PN of Δ PQR (see Fig. 7.42). Show that:
(i) Δ ABM ≅ Δ PQN
To prove: Δ ABM ≅ Δ PQN
Proof:
Since AM is the median to BC,
BM = (1/2) BC
Since PN is the median to QR,
QN = (1/2) QR
Given that BC = QR,
(1/2) BC = (1/2) QR
BM = QN — (1)
In Δ ABM and Δ PQN,
AB = PQ (Given)
BM = QN [From (1)]
AM = PN (Given)
By SSS rule,
Δ ABM ≅ Δ PQN
Therefore ∠B = ∠Q — (2) [CPCT]
(ii) Δ ABC ≅ Δ PQR
To prove: Δ ABC ≅ Δ PQR
Proof:
In Δ ABC and Δ PQR,
AB = PQ (Given)
∠B = ∠Q [From (2)]
BC = QR (Given)
By SAS rule,
Δ ABC ≅ Δ PQR
Class 9 Maths Chapter 7 Exercise 7.3 Question Answer SEBA 2026-27
4. BE and CF are two equal altitudes of a Δ ABC. Using RHS rule, prove that the Δ ABC is isosceles.
To prove: Δ ABC is an isosceles Δ (i.e., AB = AC).
Proof:
In right Δs Δ BEC and Δ CFB,
∠BEC = ∠CFB = 90° (Altitudes)
BC = CB (Hypotenuse, Common)
BE = CF (Side, Given)
By RHS rule,
Δ BEC ≅ Δ CFB
Therefore ∠BCE = ∠CBF [CPCT]
∠BCA = ∠CBA
In Δ ABC, since sides opp. to equal ∠s are equal,
AB = AC
Hence, Δ ABC is an isosceles Δ.
5. ABC is an isosceles Δ with AB = AC. Draw AP ⊥ BC to show that ∠B = ∠C.
Given: Δ ABC where AB = AC and AP ⊥ BC.
To prove: ∠B = ∠C
Proof:
In right Δs Δ APB and Δ APC,
∠APB = ∠APC = 90° (AP ⊥ BC)
AB = AC (Hypotenuse, Given)
AP = AP (Side, Common)
By RHS rule,
Δ APB ≅ Δ APC
Therefore ∠B = ∠C [CPCT]
SSS and RHS Congruence Class 9 SEBA Ex 7.3 Solutions
6. If Δ ABC ≅ Δ PQR and Δ ABC is not ≅ to Δ RPQ, then which of the following is not true.
(A) BC = PQ
(B) AC = PR
(C) QR = BC
(D) AB = PQ
Given: Δ ABC ≅ Δ PQR
Corresponding parts are equal:
AB = PQ
BC = QR
AC = PR
Now evaluating the options:
(A) BC = PQ -> Incorrect relationship (Not true)
(B) AC = PR -> True
(C) QR = BC -> True
(D) AB = PQ -> True
Answer: (A) BC = PQ
7. If two sides AC and BC and median AM of Δ ABC are respectively equal to sides PR and QR and median PN of Δ PQR (Fig. 7.43), then which of the following statements is true or false.
(i) Δ ACM ≅ Δ PRN
(ii) Δ ABC ≅ Δ PQR
Options:
(A) (i) is true, (ii) is false
(B) (i) is false, (ii) is true
(C) Both (i) and (ii) are true
(D) Both (i) and (ii) are false
Given: AC = PR, BC = QR, and median AM = PN.
Since AM and PN are medians,
MC = (1/2) BC and RN = (1/2) QR
Since BC = QR,
MC = RN
In Δ ACM and Δ PRN,
AC = PR (Given)
MC = RN (Proved)
AM = PN (Given)
By SSS rule,
Δ ACM ≅ Δ PRN — (Statement i is true)
Therefore ∠C = ∠R [CPCT]
In Δ ABC and Δ PQR,
AC = PR (Given)
∠C = ∠R (Proved)
BC = QR (Given)
By SAS rule,
Δ ABC ≅ Δ PQR — (Statement ii is true)
Answer: (C) Both (i) and (ii) are true
SEBA Class 9 Maths Chapter 7 Exercise 7.3 Textbook Answers
8. Match the columns for each pair of Δs from (P) to (S) according to their ≅ criterion (See fig 7.44).
Column A | Column B
(P) Pair with 3 equal marked sides | i) ASA
(Q) Pair with 2 sides and included ∠ marked | ii) RHS
(R) Pair with 2 ∠s and included side marked | iii) SSS
(S) Pair of right Δs with hypotenuse and one side marked | iv) SAS
Options:
A) P -> (i), Q -> (ii), R -> (iii), S -> (iv)
B) P -> (iv), Q -> (i), R -> (ii), S -> (iii)
C) P -> (iii), Q -> (iv), R -> (i), S -> (ii)
D) P -> (ii), Q -> (iv), R -> (i), S -> (iii)
Matching pairs according to figures:
(P) Three side pairs marked equal -> iii) SSS
(Q) Two sides and included ∠ marked equal -> iv) SAS
(R) Two ∠s and included side marked equal -> i) ASA
(S) Right ∠, hypotenuse, and one side marked equal -> ii) RHS
Correct sequence:
P -> (iii), Q -> (iv), R -> (i), S -> (ii)
Answer: (C) P -> (iii), Q -> (iv), R -> (i), S -> (ii)
9. A foldable step ladder of Aluminium is shown in the figure (Fig. 7.45). The length of two legs, AB and AC are both equal to 90 cm and the ∠ between the two legs as shown in the figure is 50°.
On the basis of the above information answer the following questions.
(i) ∠ACB is equal to
(A) 60°
(B) 65°
(C) 70°
(D) 75°
Given: AB = AC = 90 cm, ∠BAC = 50°.
To find ∠ACB:
In Δ ABC,
AB = AC
Therefore ∠ABC = ∠ACB (∠s opp. to equal sides are equal)
Sum of ∠s in Δ ABC = 180°:
∠BAC + ∠ABC + ∠ACB = 180°
50° + ∠ACB + ∠ACB = 180°
2∠ACB = 180° – 50°
2∠ACB = 130°
∠ACB = 65°
Answer: (B) 65°
(ii) If ∠CAB = 60° then BC =
(A) 75 cm
(B) 80 cm
(C) 85 cm
(D) 90 cm
If ∠CAB = 60° then BC =
If ∠CAB = 60° and AB = AC = 90 cm, then:
∠ABC = ∠ACB = (180° – 60°)/2 = 60°
Since all ∠s are 60°, Δ ABC is an equilateral Δ.
Therefore BC = AB = AC = 90 cm
Answer: (D) 90 cm
(iii) Δ ACB is
(A) Equilateral Δ
(B) Isosceles Δ
(C) Scalene Δ
(D) Right Δ
Type of Δ ACB:
Since AB = AC = 90 cm, the Δ has two equal sides.
Answer: (B) Isosceles Δ
(iv) In two Δs ABC and DEF, if ∠A = ∠D, AB = DE and AC = DF, then the criterion by which the Δs are ≅ is
(A) SSS
(B) SAS
(C) ASA
(D) RHS
Criterion for ≅:
Given ∠A = ∠D, AB = DE, and AC = DF.
Two sides and their included ∠ are equal.
Answer: (B) SAS
Class 9 Maths Exercise 7.3 Triangles Proofs
📐 Updated Solutions Notice: This page features complete, step-by-step SEBA Class 9 Mathematics Chapter 7 (Triangles) Exercise 7.3 Textual Solutions updated for the 2026–27 academic session. All SSS and RHS congruence criteria applications, altitude proofs, and bisector derivations strictly follow the latest revised Assam Board (SEBA) curriculum.
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