NCERT Solutions For Class 9 Maths Chapter 7 Triangles Exercise 7.3 (2025-26)

Exercise 7.3

1. \[\mathbf{\text{ }\!\!\Delta\!\!\text{ ABC}}\] and \[\mathbf{\text{ }\!\!\Delta\!\!\text{ DBC}}\] are two isosceles triangles on the same base \[\mathbf{\text{BC}}\] and vertices \[\mathbf{\text{A}}\] and \[\mathbf{\text{D}}\] are on the same side of \[\mathbf{\text{BC}}\] (see the given figure). If \[\mathbf{\text{AD}}\] is extended to intersect \[\mathbf{\text{BC}}\] at \[\mathbf{\text{P}}\], show that:

(i) \[\mathbf{\text{ }\!\!\Delta\!\!\text{ ABD}\cong \text{ }\!\!\Delta\!\!\text{ ACD}}\]

Ans: In \[\text{ }\!\!\Delta\!\!\text{ ABD}\] and \[\text{ }\!\!\Delta\!\!\text{ ACD}\],

\[\text{AB=AC}\]                                           (Given)

\[\text{BD=CD}\]                                           (Given)

\[\text{AD=AD}\]                                          (Common)

Therefore, \[\text{ }\!\!\Delta\!\!\text{ ABD}\cong \text{ }\!\!\Delta\!\!\text{ ACD}\]         (By \[\text{SSS}\] rule)

\[\angle \text{BAD=}\angle \text{CAD}\]                            (By \[CPCT\])

\[\angle \text{BAP=}\angle \text{CAP}\]

(ii) \[\mathbf{\text{ }\!\!\Delta\!\!\text{ ABP }\cong \text{ }\!\!\Delta\!\!\text{ ACP}}\]

Ans: In \[\text{ }\!\!\Delta\!\!\text{ ABP}\] and \[\text{ }\!\!\Delta\!\!\text{ ACP}\],

\[\text{AB=AC}\]                                         (Given)

\[\angle \text{BAP = }\angle \text{CAP}\]

\[\text{AP= AP}\]                                         (Common)

Therefore, \[\text{ }\!\!\Delta\!\!\text{ ABP }\cong \text{  }\!\!\Delta\!\!\text{ ACP}\]          (By \[\text{SAS}\] congruence rule)

Therefore, \[\text{BP=CP}\]                         (By \[CPCT\])

(iii) \[\mathbf{\text{AP}}\] bisects \[\mathbf{\angle \text{A}}\] as well as \[\mathbf{\angle \text{D}}\].

Ans: Since, \[\angle \text{BAP=}\angle \text{CAP}\],

Therefore, \[\text{AP}\] bisects \[\angle \text{A}\].

In \[\text{ }\!\!\Delta\!\!\text{ BDP}\] and \[\text{ }\!\!\Delta\!\!\text{ CDP}\],

\[\text{BD= CD}\]                                      (Given)

\[\text{DP=DP}\]                                        (Common)

\[\text{BP= CP}\]

Therefore, \[\text{ }\!\!\Delta\!\!\text{ BDP}\cong \text{ }\!\!\Delta\!\!\text{ CDP}\]       (By \[\text{SSS}\] congruence rule)

Therefore, \[\angle \text{BDP=}\angle \text{CDP}\]        (By \[\text{CPCT}\])

Hence, \[\text{AP}\] bisects \[\angle \text{D}\].

(iv) \[\mathbf{\text{AP}}\] is the perpendicular bisector of \[\mathbf{\text{BC}}\].

Ans: Since \[\text{ }\!\!\Delta\!\!\text{ BDP}\cong \text{ }\!\!\Delta\!\!\text{ CDP}\],

Therefore, 

\[\angle \text{BPD =}\angle \text{CPD}\]                           (By \[\text{CPCT}\])

\[\angle \text{BPD+}\angle \text{CPD=18}{{\text{0}}^{\text{o}}}\]                   (Linear pair of angles)

\[\angle \text{BPD +}\angle \text{BPD = 18}{{\text{0}}^{\text{o}}}\]                (Since \[\angle \text{BDP=}\angle \text{CDP}\])

\[\text{2}\angle \text{BPD=18}{{\text{0}}^{\text{o}}}\]

\[\angle \text{BPD = 9}{{\text{0}}^{\text{o}}}\]

Hence \[\text{AP}\] is the perpendicular bisector of \[\text{BC}\].

2. \[\mathbf{AD}\] is an altitude of an isosceles triangles \[\mathbf{ABC}\] in which \[\mathbf{\text{AB=AC}}\]. Show that:

(i) \[\mathbf{AD}\] bisects \[\mathbf{\text{BC}}\]

Ans: In \[\text{ }\!\!\Delta\!\!\text{ BAD}\] and \[\text{ }\!\!\Delta\!\!\text{ CAD}\],

\[\angle \text{ADB= }\angle \text{ADC}\]                           (Each \[\text{90 }\!\!{}^\text{o}\!\!\text{ }\])

\[\text{AB=AC}\]                                        (Given)

\[\text{AD=AD}\]                                       (Common)

Therefore, \[\text{ }\!\!\Delta\!\!\text{ BAD }\cong \text{ }\!\!\Delta\!\!\text{ CAD}\]        (By \[\text{RHS}\] congruence rule)

Therefore, \[\text{BD = CD}\]                   (By \[\text{CPCT}\])

Hence, \[\text{AD}\] bisects \[\text{BC}\].

(ii) \[\mathbf{AD}\] bisects \[\mathbf{\angle \text{A}}\].

Ans: Also, \[\angle \text{BAD=}\angle \text{CAD}\]           (By \[\text{CPCT}\])

Hence, \[AD\] bisects \[BC\].

Therefore, \[AD\] bisects \[\angle A\].

3. Two sides \[\mathbf{\text{AB}}\] and \[\mathbf{\text{BC}}\] and median \[\mathbf{\text{AM}}\] of one triangle \[\mathbf{\text{ABC}}\] are respectively equal to sides \[\mathbf{\text{PQ}}\] and \[\mathbf{\text{QR}}\] and median \[\mathbf{\text{PN}}\] of \[\mathbf{\text{ }\!\!\Delta\!\!\text{ PQR}}\] (see the given figure). Show that:

(i) \[\mathbf{\text{ }\!\!\Delta\!\!\text{ ABM}\cong \text{ }\!\!\Delta\!\!\text{ PQN}}\]

Ans: In \[\text{ }\!\!\Delta\!\!\text{ ABC}\], \[\text{AM}\] is the median to \[\text{BC}\].

Therefore, \[\text{BM=}\frac{1}{2}BC\]

In \[\text{ }\!\!\Delta\!\!\text{ PQR}\], \[PN\] is the median to \[QR\].

Therefore, \[\text{QN=}\frac{1}{2}QR\]

However, \[\text{BC=QR}\]

So, \[\text{BM=QN}\]           …… (i)

In \[\text{ }\!\!\Delta\!\!\text{ ABM}\] and \[\text{ }\!\!\Delta\!\!\text{ PQN}\],

\[\text{AB=PQ}\]                                      (Given)

\[\text{BM=QN}\]                                    (From (i))

\[\text{AM=PN}\]                                     (Given)

Therefore, \[\text{ }\!\!\Delta\!\!\text{ ABM}\cong \text{ }\!\!\Delta\!\!\text{ PQN}\]       (By \[SSS\] congruence rule)

\[\angle ABM\text{=}\angle \text{PQN}\]                         (By \[\text{CPCT}\])

So, \[\angle ABC\text{=}\angle \text{PQR}\]              …… (ii)

(ii) \[\mathbf{\text{ }\!\!\Delta\!\!\text{ ABC}\cong \text{ }\!\!\Delta\!\!\text{ PQR}}\]

Ans: In \[\text{ }\!\!\Delta\!\!\text{ ABC}\] and \[\text{ }\!\!\Delta\!\!\text{ PQR}\].

\[\text{AB=PQ}\]                                      (Given)

\[\angle ABC\text{=}\angle \text{PQR}\]                        (from (ii))

\[\text{BC=QR}\]                                   (Given)

Therefore, \[\text{ }\!\!\Delta\!\!\text{ ABC}\cong \text{ }\!\!\Delta\!\!\text{ PQR}\]     (By \[SAS\] congruence rule)

4. \[\mathbf{BE}\] and \[\mathbf{CF}\] are two equal altitudes of a triangle \[\mathbf{ABC}\]. Using \[\mathbf{RHS}\] congruence rule, prove that the triangle \[\mathbf{ABC}\] is isosceles.

Ans: In \[\text{ }\!\!\Delta\!\!\text{ BEC}\] and \[\text{ }\!\!\Delta\!\!\text{ CFB}\],

\[\angle \text{BEC= }\angle \text{CFB}\]                           (Each \[\text{90 }\!\!{}^\text{o}\!\!\text{ }\])

\[\text{BC=CB}\]                                       (Common)

\[\text{BE= CF}\]                                       (Given)

Therefore, \[\text{ }\!\!\Delta\!\!\text{ BEC}\cong \text{ }\!\!\Delta\!\!\text{ CFB}\]          (By \[RHS\] congruency)

Therefore, \[\angle \text{BCE=}\angle \text{CBF}\]           (By \[\text{CPCT}\])

Therefore, \[\text{AB=AC}\]                     (Sides opposite to equal angles of a triangle are equal)

Hence, \[\text{ }\!\!\Delta\!\!\text{ ABC}\] is isosceles.

5. \[\mathbf{ABC}\] is an isosceles triangle with \[\mathbf{\text{AB=AC}}\]. Draw \[\mathbf{\text{AP}\bot \text{BC}}\] to show that \[\mathbf{\angle \text{B=}\angle \text{C}}\].

Ans: In \[\text{ }\!\!\Delta\!\!\text{ APB}\] and \[\text{ }\!\!\Delta\!\!\text{ APC}\],

\[\angle \text{APB = }\angle \text{APC}\]                         (Each \[\text{90 }\!\!{}^\text{o}\!\!\text{ }\])

\[\text{AB=AC}\]                                      (Given)

\[\text{AP=AP}\]                                       (Common)

Therefore, \[\text{ }\!\!\Delta\!\!\text{ APB}\cong \text{ }\!\!\Delta\!\!\text{ APC}\]         (Using \[\text{RHS}\] congruence rule)

Therefore, \[\angle \text{B=}\angle \text{C}\]                    (By \[\text{CPCT}\])

Conclusion

In Class 9 Maths Chapter 7.3, we explored the various properties and theorems related to triangles. This exercise emphasized the application of the Angle Sum Property and the Exterior Angle Property of triangles. By solving the problems in 7.3 class 9 maths, we strengthened our understanding of how to find unknown angles and sides using these fundamental properties.

Class 9 Maths Chapter 7: Exercises Breakdown

CBSE Class 9 Maths Chapter 7 Other Study Materials

NCERT Solutions for Class 9 Maths - Chapter-wise PDF

Given below are the chapter-wise NCERT Solutions for Class 9 Maths. Go through these  9th maths solutions chapter-wise to be thoroughly familiar with the concepts.

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