NCERT Solutions for Class 9 Maths Chapter 7 Exercise 7.2 - Triangles

Exercise 7.2

1. In an isosceles triangle $\text{ABC}$, with $\text{AB}=\text{AC}$, the bisectors of $\angle \text{B}$ and $\angle \text{C}$ intersect each other at $O$. Join $\text{A}$ to $\text{O}$. Show that:

(i) $\text{OB}=\text{OC}$

(ii) $AO$ bisects $\angle \text{A}$

Ans:

(i) It is given that in triangle $\text{ABC}$ is an isosceles triangle with $\text{AB}=\text{AC}$.

$\therefore \,\,\angle \text{ACB}=\angle \text{ABC}$ (Angles opposite to equal sides of a triangle are equal) 

Multiplying $\frac{1}{2}$ on both sides, we get

$\frac{1}{2}\angle \text{ACB}=\frac{1}{2}\angle \text{ABC}$

$\angle \text{OCB}=\angle \text{OBC}$

$\therefore \text{OB}=\text{OC}$ (Sides opposite to equal angles of a triangle are also equal)

(ii) In $\Delta \text{OAB}$ and $\Delta \text{OAC}$,

Side AO is present in both triangle. Therefore,

$\text{AO}=\text{AO}$ (Common)

Length of side AB is equal to AC, i.e., $\text{AB}=\text{AC}$ (Given)

$\text{OB}=\text{OC}$ (Proved above)

Therefore,

$\Delta \text{OAB}\cong \Delta \text{OAC}$ (By SSS axiom of congruency)

Therefore, $\angle \text{BAO}=\angle \text{CAO}$.

As they are corresponding parts of the corresponding triangle.

$\therefore $ AO bisects $\angle \text{A}$.     ($\angle \text{BAO}=\angle \text{CAO}$)

2. In $\Delta \text{ABC},\text{AD}$ is the perpendicular bisector of $\text{BC}$ (see the given figure). Show that $\vartriangle \text{ABC}$ is an isosceles triangle in which $\text{AB}=\text{AC}$.

Ans:

In $\Delta \text{ADC}$ and $\Delta \text{ADB}$,

Side AD is present in both triangle. 

$\therefore \,\,\text{AD}=\text{AD}$ (Common)

Since AD is the perpendicular bisector of BC.

$\angle ADC=\angle ADB={{90}^{{}^\circ }}$ and

$\text{CD}=\text{BD}$

$\therefore \Delta \text{ADC}\cong \Delta \text{ADB}$ (By SAS axiom of congruency)

Therefore, $\text{AB}=\text{AC}$.

As they are corresponding parts of the corresponding triangle.

Hence, $\Delta ABC$ is an isosceles triangle as two of it’s sides are equal, i.e., $AB=AC$.

3. $\text{ABC}$ is an isosceles triangle in which altitudes $\text{BE}$ and $\text{CF}$ are drawn to equal sides $\text{AC}$ and $\text{AB}$ respectively (see the given figure). Show that these altitudes are equal.

Ans:

In $\Delta \text{AEB}$ and $\Delta \text{AFC}$,

Since $\text{BE}$ and $\text{CF}$ are drawn to equal sides $\text{AC}$ and $\text{AB}$. Therefore,

$\angle \text{AEB}$ and $\angle \text{AFC}$ (Each \[{{90}^{{}^\circ }}\])

As $\angle A$ lies in both the triangles. Therefore,

$\angle \text{A}=\angle \text{A}$ (Common angle)

Length of side AB is equal to AC, i.e.,$\text{AB}=\text{AC}$ (Given)

$\therefore \Delta \text{AEB}\cong \Delta \text{AFC}$ (By AAS axiom of congruency) 

Therefore, $\text{BE}=\text{CF}$.

As they are corresponding parts of the corresponding triangle.

4.$\text{ABC}$ is a triangle in which altitudes $\text{BE}$ and $\text{CF}$ to sides $\text{AC}$ and $\text{AB}$ are equal (see the given figure). Show that

(i) $\Delta \text{ABE}\cong \Delta \text{ACF}$

(ii) $\text{AB}=\text{AC}$, ie., $\text{ABC}$ is an isosceles triangle.

Ans:

(i) In $\Delta \text{ABE}$ and $\Delta \text{ACF}$,

$\text{ABC}$ is a triangle in which altitudes $\text{BE}$ and $\text{CF}$ to sides $\text{AC}$ and $\text{AB}$.

Therefore, $\angle \text{ABE}$ and $\angle ACF={{90}^{{}^\circ }}$

Since $\angle A$ is present in both triangles,

$\therefore \,\,\angle \text{A}=\angle \text{A}$ (Common angle)

Length of side BE is equal to CF, i.e., $\text{BE}=\text{CF}$ (Given)

$\therefore \Delta \text{ABE}\cong \Delta \text{ACF}$ (By AAS axiom of congruency)

(ii) Since $\Delta \text{ABE}\cong \Delta \text{ACF}$   (proved above)

Therefore, $\text{AB}=\text{AC}$.

As they are corresponding parts of the corresponding triangle.

5. $\text{ABC}$ and $\text{DBC}$ are two isosceles triangles on the same base $\text{BC}$ (see the given figure). Show that $\angle \text{ABD}=\angle \text{ACD}$.

Ans:

Construction: join $\text{AD}$.

In $\Delta \text{ABD}$ and $\Delta \text{ACD}$,

Length of side AB is equal to AC, i.e., $\text{AB}=\text{AC}$ (Given)

Length of side BD is equal to CD, i.e., $\text{BD}=\text{CD}$ (Given)

As side AD is present in both triangles. Therefore,

$\text{AD}=\text{AD}$ (Common side)

$\therefore \Delta \text{ABD}\cong \Delta \text{ACD}$ (By SSS axiom of congruency)

Therefore, $\angle \text{ABD}=\angle \text{ACD}$.

As they are corresponding parts of the corresponding triangle.

6. $\Delta \text{ABC}$ is an isosceles triangle in which $\text{AB}=\text{AC}$. Side $\text{BA}$ is produced to $\text{D}$ such that $\text{AD}=\text{AB}$ (see the given figure). Show that $\angle \text{BCD}$ is a right angle.

Ans:

In $\Delta \text{ABC}$,

Length of side AB is equal to AC, i.e., $\text{AB}=\text{AC}$ (Given)

$\therefore \angle \text{ACB}=\angle \text{ABC}$ (Angles opposite to equal sides of a triangle are also equal)

In $\Delta \text{ACD}$,

Length of side AC is equal to AD, i.e., $\text{AC}=\text{AD}$  (Given)

$\therefore \angle \text{ADC}=\angle \text{ACD}$ (Angles opposite to equal sides of a triangle are also equal)

In $\Delta \text{BCD}$,

$\angle \text{ABC}+\angle \text{BCD}+\angle \text{ADC}={{180}^{{}^\circ }}$ (Sum of angles in a triangle is ${{180}^{{}^\circ }}$)

$\angle \text{ACB}+\angle \text{ACB}+\angle \text{ACD}+\angle \text{ACD}={{180}^{{}^\circ }}$  ($\angle BCD=\angle BCA+\angle ACD$)

$2(\angle \text{ACB}+\angle \text{ACD})={{180}^{{}^\circ }}$

$2(\angle \text{BCD})={{180}^{{}^\circ }}$

$\therefore \angle \text{BCD}={{90}^{{}^\circ }}$

7. $\text{ABC}$ is a right-angled triangle in which $\angle \text{A}={{90}^{{}^\circ }}$ and $\text{AB}=\text{AC}$. Find $\angle \text{B}$ and $\angle \text{C}$.

Ans:

It is given that the side $\text{AB}=\text{AC}$

Angles opposite to equal sides are also equal

$\therefore \angle \text{C}=\angle \text{B}$  …(i)

In $\Delta \text{ABC}$,

$\angle \text{A}+\angle \text{B}+\angle \text{C}={{180}^{{}^\circ }}$ (Sum of angles in a triangle is ${{180}^{{}^\circ }}$)${{90}^{{}^\circ }}+\angle \text{B}+\angle \text{C}={{180}^{{}^\circ }}$  …from equation (i)

${{90}^{{}^\circ }}+\angle \text{B}+\angle \text{B}={{180}^{{}^\circ }}$

$2\angle \text{B}={{90}^{{}^\circ }}$

$\angle \text{B}={{45}^{{}^\circ }}$

$\begin{align} & \because \angle \text{B}=\angle \text{C}\,\,\,\,\,\,\,\,\,\,\,...\text{from equation}\,\left( i \right) \\  & \therefore \angle \text{C}={{45}^{{}^\circ }} \\ \end{align}$

8. Show that the angles of an equilateral triangle are ${{60}^{{}^\circ }}$ each.

Ans:

Consider that $\text{ABC}$ is an equilateral triangle in which all sides are equal.

Therefore, $\text{AB}=\text{BC}=\text{AC}$

$\text{AB}=\text{AC}$

Angles opposite to equal sides of a triangle are equal

$\therefore \angle \text{C}=\angle \text{B}\,\,\,\,\,\,...\left( i \right)$ 

Also, $\text{AC}=\text{BC}$

Angles opposite to equal sides of a triangle are equal

$\therefore \angle \text{B}=\angle \text{A}\,\,\,\,\,\,\,\,\,...\left( ii \right)$ 

Therefore, from equation (i) and (ii), we get

$\angle \text{A}=\angle \text{B}=\angle \text{C}\,\,\,\,\,\,\,\,...\left( iii \right)$

In $\Delta \text{ABC}$,

$\angle \text{A}+\angle \text{B}+\angle \text{C}={{180}^{{}^\circ }}$   (Sum of angles in a triangle is ${{180}^{{}^\circ }}$)

$\angle \text{A}+\angle \text{A}+\angle \text{A}={{180}^{{}^\circ }}$     …from equation (iii)

$3\angle \text{A}={{180}^{{}^\circ }}$

$\angle \text{A}={{60}^{{}^\circ }}$

$\therefore \angle \text{A}=\angle \text{B}=\angle \text{C}={{60}^{{}^\circ }}$

From above we can conclude that all the interior angles of an equilateral triangle has an angle of ${{60}^{{}^\circ }}$.

Conclusion

In Class 9 Maths Ch 7 Ex 7.2 it is important to understand and focus on the properties and criteria for triangle congruence, such as SAS (Side-Angle-Side), ASA (Angle-Side-Angle), and SSS (Side-Side-Side). These concepts are crucial for solving problems related to triangle congruence. Pay attention to how these criteria are applied to different questions to prove that two triangles are congruent. Understanding these principles will help you tackle various geometric problems and establish a solid foundation for future studies in geometry.

Class 9 Maths Chapter 7: Exercises Breakdown

CBSE Class 9 Maths Chapter 7 Other Study Materials

Other than Maths Class 9 Chapter 7 Exercise 7.2  you can also check on the additional study materials provided for Class 9 Maths Chapter 7.

Chapter-Specific NCERT Solutions for Class 9 Maths

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