Exercise 7.2 Practice
Properties of Isosceles Triangles
Q1: Isosceles Bisectors
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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) $OB = OC$
(ii) $AO$ bisects $\angle A$
(i) OB = OC:
Since $AB = AC$, $\angle B = \angle C$ (Angles opposite to equal sides).
$\Rightarrow \frac{1}{2}\angle B = \frac{1}{2}\angle C$.
Since $OB$ and $OC$ are bisectors, $\angle OBC = \angle OCB$.
In $\triangle OBC$, sides opposite to equal angles are equal.
$\therefore OB = OC$.
Since $AB = AC$, $\angle B = \angle C$ (Angles opposite to equal sides).
$\Rightarrow \frac{1}{2}\angle B = \frac{1}{2}\angle C$.
Since $OB$ and $OC$ are bisectors, $\angle OBC = \angle OCB$.
In $\triangle OBC$, sides opposite to equal angles are equal.
$\therefore OB = OC$.
(ii) AO bisects $\angle A$:
In $\triangle ABO$ and $\triangle ACO$:
1. $AB = AC$ (Given)
2. $OB = OC$ (Proved above)
3. $AO = AO$ (Common)
$\therefore \triangle ABO \cong \triangle ACO$ (SSS Rule).
$\Rightarrow \angle BAO = \angle CAO$ (CPCT).
Thus, $AO$ bisects $\angle A$.
In $\triangle ABO$ and $\triangle ACO$:
1. $AB = AC$ (Given)
2. $OB = OC$ (Proved above)
3. $AO = AO$ (Common)
$\therefore \triangle ABO \cong \triangle ACO$ (SSS Rule).
$\Rightarrow \angle BAO = \angle CAO$ (CPCT).
Thus, $AO$ bisects $\angle A$.
Proved.
Q2: Perpendicular Bisector
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In $\triangle ABC$, $AD$ is the perpendicular bisector of $BC$. Show that $\triangle ABC$ is an isosceles triangle in which $AB = AC$.
In $\triangle ABD$ and $\triangle ACD$:
1. $BD = CD$ (AD is bisector)
2. $\angle ADB = \angle ADC = 90^\circ$ (AD is perpendicular)
3. $AD = AD$ (Common)
1. $BD = CD$ (AD is bisector)
2. $\angle ADB = \angle ADC = 90^\circ$ (AD is perpendicular)
3. $AD = AD$ (Common)
$\therefore \triangle ABD \cong \triangle ACD$ (SAS Rule).
By CPCT, $AB = AC$.
Thus, $\triangle ABC$ is isosceles.
Thus, $\triangle ABC$ is isosceles.
Proved.
Q3: Altitudes to Equal Sides
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$ABC$ 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.
In $\triangle ABE$ and $\triangle ACF$:
1. $\angle AEB = \angle AFC = 90^\circ$ (Altitudes)
2. $\angle A = \angle A$ (Common angle)
3. $AB = AC$ (Given)
1. $\angle AEB = \angle AFC = 90^\circ$ (Altitudes)
2. $\angle A = \angle A$ (Common angle)
3. $AB = AC$ (Given)
$\therefore \triangle ABE \cong \triangle ACF$ (AAS Rule).
By CPCT, $BE = CF$.
Proved.
Q4: Equal Altitudes
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$ABC$ is a triangle in which altitudes $BE$ and $CF$ to sides $AC$ and $AB$ are equal. Show that:
(i) $\triangle ABE \cong \triangle ACF$
(ii) $AB = AC$, i.e., $ABC$ is an isosceles triangle.
(i) $\triangle ABE \cong \triangle ACF$
(ii) $AB = AC$, i.e., $ABC$ is an isosceles triangle.
(i) In $\triangle ABE$ and $\triangle ACF$:
1. $\angle AEB = \angle AFC = 90^\circ$
2. $\angle A = \angle A$ (Common)
3. $BE = CF$ (Given)
$\therefore \triangle ABE \cong \triangle ACF$ (AAS Rule).
1. $\angle AEB = \angle AFC = 90^\circ$
2. $\angle A = \angle A$ (Common)
3. $BE = CF$ (Given)
$\therefore \triangle ABE \cong \triangle ACF$ (AAS Rule).
(ii) AB = AC:
By CPCT, $AB = AC$.
Thus, $\triangle ABC$ is isosceles.
By CPCT, $AB = AC$.
Thus, $\triangle ABC$ is isosceles.
Proved.
Q5: Two Isosceles Triangles
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$ABC$ and $DBC$ are two isosceles triangles on the same base $BC$. Show that $\angle ABD = \angle ACD$.
Since $\triangle ABC$ is isosceles ($AB = AC$), $\angle ABC = \angle ACB$.
Since $\triangle DBC$ is isosceles ($DB = DC$), $\angle DBC = \angle DCB$.
Adding the two equations:
$\angle ABC + \angle DBC = \angle ACB + \angle DCB$
$\Rightarrow \angle ABD = \angle ACD$.
$\angle ABC + \angle DBC = \angle ACB + \angle DCB$
$\Rightarrow \angle ABD = \angle ACD$.
Proved.
Q6: Produced Side
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$\triangle ABC$ 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.
In $\triangle ABC$, $AB = AC \Rightarrow \angle ACB = \angle ABC$. Let this be $x$.
In $\triangle ACD$, $AD = AB$ (Given) and $AB = AC$. So $AD = AC$.
$\Rightarrow \angle ACD = \angle ADC$. Let this be $y$.
$\Rightarrow \angle ACD = \angle ADC$. Let this be $y$.
In $\triangle BCD$, sum of angles is $180^\circ$:
$\angle B + \angle C + \angle D = 180^\circ$
$x + (x + y) + y = 180^\circ$
$2(x + y) = 180^\circ \Rightarrow x + y = 90^\circ$.
$\angle B + \angle C + \angle D = 180^\circ$
$x + (x + y) + y = 180^\circ$
$2(x + y) = 180^\circ \Rightarrow x + y = 90^\circ$.
$\angle BCD = x + y = 90^\circ$.
Proved.
Q7: Right Isosceles
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$ABC$ is a right angled triangle in which $\angle A = 90^\circ$ and $AB = AC$. Find $\angle B$ and $\angle C$.
Since $AB = AC$, $\angle B = \angle C$.
In $\triangle ABC$:
$\angle A + \angle B + \angle C = 180^\circ$
$90^\circ + \angle B + \angle B = 180^\circ$
$2\angle B = 90^\circ \Rightarrow \angle B = 45^\circ$.
$\angle A + \angle B + \angle C = 180^\circ$
$90^\circ + \angle B + \angle B = 180^\circ$
$2\angle B = 90^\circ \Rightarrow \angle B = 45^\circ$.
$\angle B = 45^\circ, \angle C = 45^\circ$
Q8: Equilateral Angles
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Show that the angles of an equilateral triangle are $60^\circ$ each.
Let $\triangle ABC$ be equilateral.
$AB = BC = AC$.
$AB = BC = AC$.
Since $AB = AC$, $\angle C = \angle B$.
Since $AC = BC$, $\angle B = \angle A$.
$\therefore \angle A = \angle B = \angle C$.
Since $AC = BC$, $\angle B = \angle A$.
$\therefore \angle A = \angle B = \angle C$.
$\angle A + \angle B + \angle C = 180^\circ$
$3\angle A = 180^\circ \Rightarrow \angle A = 60^\circ$.
$3\angle A = 180^\circ \Rightarrow \angle A = 60^\circ$.
Each angle is $60^\circ$.