Triangles
NCERT Solutions • Class 9 Maths • Chapter 7Exercise 7.2
1. In an isosceles triangle ABC, with $AB = AC$, the bisectors of $\angle B$ and $\angle C$ intersect each other at $O$. Join $A$ to $O$. Show that: (i) $OB = OC$ (ii) $AO$ bisects $\angle A$.
(i) Show $OB = OC$:
Since $AB = AC$, $\angle B = \angle C$ (Angles opposite to equal sides).
Dividing both sides by 2: $\frac{1}{2}\angle B = \frac{1}{2}\angle C$.
Since $OB$ and $OC$ are bisectors, $\angle OBC = \angle OCB$.
In $\Delta OBC$, sides opposite to equal angles are equal.
$\therefore$ $OB = OC$.
Since $AB = AC$, $\angle B = \angle C$ (Angles opposite to equal sides).
Dividing both sides by 2: $\frac{1}{2}\angle B = \frac{1}{2}\angle C$.
Since $OB$ and $OC$ are bisectors, $\angle OBC = \angle OCB$.
In $\Delta OBC$, sides opposite to equal angles are equal.
$\therefore$ $OB = OC$.
(ii) Show $AO$ bisects $\angle A$:
In $\Delta ABO$ and $\Delta ACO$:
1. $AB = AC$ (Given)
2. $AO = AO$ (Common)
3. $OB = OC$ (Proved above)
$\therefore \Delta ABO \cong \Delta ACO$ (By SSS Rule).
$\Rightarrow \angle BAO = \angle CAO$ (CPCT).
Hence, $AO$ bisects $\angle A$.
In $\Delta ABO$ and $\Delta ACO$:
1. $AB = AC$ (Given)
2. $AO = AO$ (Common)
3. $OB = OC$ (Proved above)
$\therefore \Delta ABO \cong \Delta ACO$ (By SSS Rule).
$\Rightarrow \angle BAO = \angle CAO$ (CPCT).
Hence, $AO$ bisects $\angle A$.
2. In $\Delta ABC$, $AD$ is the perpendicular bisector of $BC$. Show that $\Delta ABC$ is an isosceles triangle in which $AB = AC$.
Given: $AD \perp BC$ and $BD = CD$.
Proof: In $\Delta ABD$ and $\Delta ACD$:
1. $AD = AD$ (Common)
2. $\angle ADB = \angle ADC = 90^\circ$ (Given)
3. $BD = CD$ ($AD$ is bisector)
$\therefore \Delta ABD \cong \Delta ACD$ (By SAS Rule).
$\Rightarrow$ $AB = AC$ (CPCT).
Thus, $\Delta ABC$ is an isosceles triangle.
Proof: In $\Delta ABD$ and $\Delta ACD$:
1. $AD = AD$ (Common)
2. $\angle ADB = \angle ADC = 90^\circ$ (Given)
3. $BD = CD$ ($AD$ is bisector)
$\therefore \Delta ABD \cong \Delta ACD$ (By SAS Rule).
$\Rightarrow$ $AB = AC$ (CPCT).
Thus, $\Delta ABC$ is an isosceles triangle.
3. $ABC$ is an isosceles triangle in which altitudes $BE$ and $CF$ are drawn to equal sides $AC$ and $AB$ respectively. Show that these altitudes are equal.
Proof: In $\Delta ABE$ and $\Delta ACF$:
1. $\angle A = \angle A$ (Common angle)
2. $\angle AEB = \angle AFC = 90^\circ$ (Altitudes)
3. $AB = AC$ (Given)
$\therefore \Delta ABE \cong \Delta ACF$ (By AAS Rule).
$\Rightarrow$ $BE = CF$ (CPCT).
1. $\angle A = \angle A$ (Common angle)
2. $\angle AEB = \angle AFC = 90^\circ$ (Altitudes)
3. $AB = AC$ (Given)
$\therefore \Delta ABE \cong \Delta ACF$ (By AAS Rule).
$\Rightarrow$ $BE = CF$ (CPCT).
4. $ABC$ is a triangle in which altitudes $BE$ and $CF$ to sides $AC$ and $AB$ are equal. Show that:
(i) $\Delta ABE \cong \Delta ACF$
(ii) $AB = AC$.
(i) $\Delta ABE \cong \Delta ACF$
(ii) $AB = AC$.
(i) In $\Delta ABE$ and $\Delta ACF$:
1. $\angle A = \angle A$ (Common)
2. $\angle AEB = \angle AFC = 90^\circ$
3. $BE = CF$ (Given)
$\therefore$ $\Delta ABE \cong \Delta ACF$ (By AAS Rule).
1. $\angle A = \angle A$ (Common)
2. $\angle AEB = \angle AFC = 90^\circ$
3. $BE = CF$ (Given)
$\therefore$ $\Delta ABE \cong \Delta ACF$ (By AAS Rule).
(ii) From the congruence above:
$AB = AC$ (CPCT).
Hence, $ABC$ is an isosceles triangle.
$AB = AC$ (CPCT).
Hence, $ABC$ is an isosceles triangle.
5. $ABC$ and $DBC$ are two isosceles triangles on the same base $BC$. Show that $\angle ABD = \angle ACD$.
Proof:
1. In $\Delta ABC$, $AB = AC \Rightarrow \angle ABC = \angle ACB$ (Angles opposite equal sides).
2. In $\Delta DBC$, $DB = DC \Rightarrow \angle DBC = \angle DCB$.
Adding both equations:
$\angle ABC + \angle DBC = \angle ACB + \angle DCB$
$\Rightarrow$ $\angle ABD = \angle ACD$.
1. In $\Delta ABC$, $AB = AC \Rightarrow \angle ABC = \angle ACB$ (Angles opposite equal sides).
2. In $\Delta DBC$, $DB = DC \Rightarrow \angle DBC = \angle DCB$.
Adding both equations:
$\angle ABC + \angle DBC = \angle ACB + \angle DCB$
$\Rightarrow$ $\angle ABD = \angle ACD$.
6. $\Delta ABC$ is an isosceles triangle in which $AB = AC$. Side $BA$ is produced to $D$ such that $AD = AB$. Show that $\angle BCD$ is a right angle.
Step 1: In $\Delta ABC$, $AB = AC \Rightarrow \angle ACB = \angle ABC$. (Let this be $\angle 1$).
Step 2: In $\Delta ACD$, $AC = AD$ (since $AB=AC$ and $AB=AD$).
$\Rightarrow \angle ACD = \angle ADC$. (Let this be $\angle 2$).
Step 2: In $\Delta ACD$, $AC = AD$ (since $AB=AC$ and $AB=AD$).
$\Rightarrow \angle ACD = \angle ADC$. (Let this be $\angle 2$).
Step 3: In $\Delta BCD$, sum of angles is $180^\circ$.
$\angle B + \angle D + \angle BCD = 180^\circ$
$\angle 1 + \angle 2 + (\angle 1 + \angle 2) = 180^\circ$
$2(\angle 1 + \angle 2) = 180^\circ$
$\angle 1 + \angle 2 = 90^\circ$
Since $\angle BCD = \angle 1 + \angle 2$, $\angle BCD = 90^\circ$.
$\angle B + \angle D + \angle BCD = 180^\circ$
$\angle 1 + \angle 2 + (\angle 1 + \angle 2) = 180^\circ$
$2(\angle 1 + \angle 2) = 180^\circ$
$\angle 1 + \angle 2 = 90^\circ$
Since $\angle BCD = \angle 1 + \angle 2$, $\angle BCD = 90^\circ$.
7. $ABC$ is a right angled triangle in which $\angle A = 90^\circ$ and $AB = AC$. Find $\angle B$ and $\angle C$.
Since $AB = AC$, angles opposite to them are equal:
$\angle B = \angle C$.
By Angle Sum Property:
$\angle A + \angle B + \angle C = 180^\circ$
$90^\circ + \angle B + \angle B = 180^\circ$
$2\angle B = 90^\circ \Rightarrow \angle B = 45^\circ$.
Therefore, $\angle B = 45^\circ$ and $\angle C = 45^\circ$.
$\angle B = \angle C$.
By Angle Sum Property:
$\angle A + \angle B + \angle C = 180^\circ$
$90^\circ + \angle B + \angle B = 180^\circ$
$2\angle B = 90^\circ \Rightarrow \angle B = 45^\circ$.
Therefore, $\angle B = 45^\circ$ and $\angle C = 45^\circ$.
8. Show that the angles of an equilateral triangle are $60^\circ$ each.
Let $\Delta ABC$ be equilateral. So, $AB = BC = AC$.
Since $AB = BC \Rightarrow \angle C = \angle A$.
Since $BC = AC \Rightarrow \angle A = \angle B$.
Thus, $\angle A = \angle B = \angle C$.
Using Angle Sum Property:
$3\angle A = 180^\circ \Rightarrow \angle A = 60^\circ$.
Hence, each angle is $60^\circ$.
Since $AB = BC \Rightarrow \angle C = \angle A$.
Since $BC = AC \Rightarrow \angle A = \angle B$.
Thus, $\angle A = \angle B = \angle C$.
Using Angle Sum Property:
$3\angle A = 180^\circ \Rightarrow \angle A = 60^\circ$.
Hence, each angle is $60^\circ$.