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

Q1 (i) In an isosceles triangle ABC, with $AB=AC$ , the bisectors of $\mathrm{\angle }B$ and $\mathrm{\angle }C$ intersect each other at O. Join A to O. Show that : $OB=OC$

Answer:

Given, AB = AC, and ABC is an isosceles triangle.

Therefore, it means $\mathrm{\angle }B=\mathrm{\angle }C$

or $\frac{1}{2}\mathrm{\angle }B=\frac{1}{2}\mathrm{\angle }C$

Thus, $\mathrm{\angle }OBC=\mathrm{\angle }OCB$ (Angles bisectors are equal)

Hence, $OB=OC$ as sides opposite to equal are angles are also equal.

Q1 (ii) In an isosceles triangle ABC, with $AB=AC$ , the bisectors of $\mathrm{\angle }B$ and $\mathrm{\angle }C$ intersect each other at O. Join A to O. Show that : AO bisects $\mathrm{\angle }A$

Answer:

Consider $\mathrm{\Delta }AOB$ and $\mathrm{\Delta }AOC$ ,

(i) $AB=AC$ (Given)

(ii) $AO=AO$ (Common in both the triangles)

(iii) $OB=OC$ (Proved in previous part)

Therefore by SSS congruence rule, we can conclude that :

$\mathrm{\Delta }AOB\cong \mathrm{\Delta }AOC$

Now, by c.p.c.t.,

$\mathrm{\angle }BAO=\mathrm{\angle }CAO$

Hence AO bisects $\mathrm{\angle }A$ .

Q2 In $\mathrm{\Delta }ABC$ , AD is the perpendicular bisector of BC (see Fig). Show that $\mathrm{\Delta }ABC$ is an isosceles triangle in which $AB=AC$ .

Answer:

Given: AD is perpendicular bisector of BC.

To Prove: AB = AC

Proof: Consider $\mathrm{\Delta }$ ABD and $\mathrm{\Delta }$ ADC,

(i) $AD=AD$ (Common in both the triangles)

(ii) $\mathrm{\angle }ADB=\mathrm{\angle }ADC$ (Right angle)

(iii) BD = CD ( as AD is perpendicular to side BC)

Therefore, by SAS congruence criteria:

$\mathrm{\Delta }ADB\cong \mathrm{\Delta }ADC$

Thus, AB = AC ( by c.p.c.t )

Hence Proved

Q3 ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively (see Fig.). Show that these altitudes are equal.

Answer:

Given: AC = AB and BE and CF are altitudes.

To Prove: BE = CF

Proof: Consider $\mathrm{\Delta }AEB$ and $\mathrm{\Delta }AFC$ ,

(i) $\mathrm{\angle }A$ is common in both the triangles.

(ii) $\mathrm{\angle }AEB=\mathrm{\angle }AFC$ (Right angles)

(iii) $AB=AC$ (Given)

Thus by AAS congruence axiom, we can conclude that :

$\mathrm{\Delta }AEB\cong \mathrm{\Delta }AFC$

Now, by c.p.c.t. we can say : $BE=CF$

Hence Proved

Q4 (i) ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig). Show that $\mathrm{\Delta }ABE\cong \mathrm{\Delta }ACF$

Answer:

Given: BE = CF

Consider $\mathrm{\Delta }ABE$ and $\mathrm{\Delta }ACF$ ,

(i) $\mathrm{\angle }A$ is common in both the triangles.

(ii) $\mathrm{\angle }AEB=\mathrm{\angle }AFC$ (Right angles)

(iii) $BE=CF$ (Given)

Thus by AAS congruence, we can say that :

$\mathrm{\Delta }ABE\cong \mathrm{\Delta }ACF$

Hence Proved

Q4 (ii) ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see Fig.). Show that $AB=AC$ , i.e., ABC is an isosceles triangle.

Answer:

From the prevoius part of the question we found out that : $\mathrm{\Delta }ABE\cong \mathrm{\Delta }ACF$

Now, by c.p.c.t. we can say that : $AB=AC$

Thus, $\mathrm{\Delta }ABC$ is an isosceles triangle.

Hence Proved

Q5 ABC and DBC are two isosceles triangles on the same base BC (see Fig.). Show that $\mathrm{\angle }ABD=\mathrm{\angle }ACD$.

Answer:

Given: ABC and DBC are isosceles triangles and

To Prove: $\mathrm{\angle }ABD=\mathrm{\angle }ACD$

Proof: Consider $\mathrm{\Delta }ABD$ and $\mathrm{\Delta }ACD$,

(i) $AD=AD$ (Common in both the triangles)

(ii) $AB=AC$ (Sides of isosceles triangle)

(iii) $BD=CD$ (Sides of isosceles triangle)

Thus by SSS congruency, we can conclude that:

$\mathrm{△}ABD\cong triangleACD$

Therefore, by c.p.c.t.,

$\mathrm{\angle }ABD=\mathrm{\angle }ACD$

Hence Proved

Q6 $\mathrm{\Delta }ABC$ is an isosceles triangle in which $AB=AC$ . Side BA is produced to D such that $AD=AB$ (see Fig.). Show that $\mathrm{\angle }BCD$ is a right angle.

Answer:

Given: AB = AC and AD = AB

To Prove: $\mathrm{\angle }BCD$ is a right angle

Proof: Consider $\mathrm{\Delta }$ ABC,
It is given that AB = AC
So, $\mathrm{\angle }ACB=\mathrm{\angle }ABC$ (Since angles opposite to the equal sides are equal.)

Similarly in $\mathrm{\Delta }$ ACD, AD = AB and $\mathrm{\angle }ADC=\mathrm{\angle }ACD$
So,

$\mathrm{\angle }CAB+\mathrm{\angle }ACB+\mathrm{\angle }ABC={180}^{\circ }$

$\mathrm{\angle }CAB+2\mathrm{\angle }ACB={180}^{\circ }$
or $\mathrm{\angle }CAB={180}^{\circ }-2\mathrm{\angle }ACB$ ...........................(i)

And in $\mathrm{\Delta }$ ADC,
$\mathrm{\angle }CAD={180}^{\circ }-2\mathrm{\angle }ACD$ ..............................(ii)

Adding (i) and (ii), we get :
$\mathrm{\angle }CAB+\mathrm{\angle }CAD={360}^{\circ }-2\mathrm{\angle }ACD-2\mathrm{\angle }ACB$

or ${180}^{\circ }={360}^{\circ }-2\mathrm{\angle }ACD-2\mathrm{\angle }ACB$

and $\mathrm{\angle }BCD={90}^{\circ }$

Hence Proved

Q7 ABC is a right angled triangle in which $\mathrm{\angle }A={90}^{\circ }$ and $AB=AC$ . Find $\mathrm{\angle }B$ and $\mathrm{\angle }C$ .

Answer:

Given: AB = AC and $\mathrm{\angle }A={90}^{\circ }$

We know that angles opposite to equal sides are also equal.

Therefore, $\mathrm{\angle }B=\mathrm{\angle }C$

Also, the sum of the interior angles of a triangle is ${180}^{\circ }$ .

So, we have :

$\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }$

or ${90}^{\circ }+2\mathrm{\angle }B={180}^{\circ }$

or $\mathrm{\angle }B={45}^{\circ }$

As, $\mathrm{\angle }B=\mathrm{\angle }C$

So, $angleC$=\ 45^{\circ}$

Q8 Show that the angles of an equilateral triangle are ${60}^{\circ }$ each.

Answer:

Consider a triangle ABC which has all sides equal as shown in the figure.

We know that angles opposite to equal sides are equal.

Therefore: $\mathrm{\angle }A=\mathrm{\angle }B=\mathrm{\angle }C$

Also, the sum of the interior angles of a triangle is ${180}^{\circ }$ .

Hence, $\mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }$

or $3\mathrm{\angle }A={180}^{\circ }$

or $\mathrm{\angle }A={60}^{\circ }$

As, all the angles of the equilateral triangle are equal, thus $\mathrm{\angle }A=\mathrm{\angle }B=\mathrm{\angle }C$ = ${60}^{\circ }$.