Important Formulas, Rules & Key Points

• Isosceles Triangle: A triangle with at least two equal sides.
• Theorem 7.2: Angles opposite to equal sides of an isosceles triangle are equal.
• Theorem 7.3: The sides opposite to equal angles of a triangle are equal.
• Angle Sum Property: The sum of all angles in a triangle is always $180^\circ$.
• Congruence Rules: Understand and apply SAS (Side-Angle-Side), ASA (Angle-Side-Angle), SSS (Side-Side-Side), and AAS (Angle-Angle-Side) criteria when solving these problems.

Q1. In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other at O. Join A to O. Show that:
(i) OB = OC
(ii) AO bisects ∠A.

Proof:

(i) To prove OB = OC:

• In $\triangle ABC$, we are given $AB = AC$.
• Therefore, $\angle C = \angle B$ (Angles opposite to equal sides are equal).
• Multiplying both sides by $\frac{1}{2}$:
  $\frac{1}{2}\angle C = \frac{1}{2}\angle B \implies \angle OCB = \angle OBC$ (Since $OB$ and $OC$ bisect $\angle B$ and $\angle C$).
• In $\triangle OBC$, since $\angle OBC = \angle OCB$, we get $OB = OC$ (Sides opposite to equal angles of a triangle are equal). [Hence Proved]

(ii) To prove AO bisects ∠A:

• In $\triangle AOB$ and $\triangle AOC$:
  • $AB = AC$ (Given)
  • $OB = OC$ (Proved above)
  • $AO = AO$ (Common side)
• By SSS congruence criterion, $\triangle AOB \cong \triangle AOC$.
• Therefore, $\angle BAO = \angle CAO$ by CPCT (Corresponding Parts of Congruent Triangles).
• This implies $AO$ bisects $\angle A$. [Hence Proved]

Q2. In ▵ABC, AD is the perpendicular bisector of BC. Show that ▵ABC is an isosceles triangle in which AB = AC.

Proof:

• In $\triangle ABD$ and $\triangle ACD$:
  • $BD = CD$ (Since $AD$ is the bisector of $BC$)
  • $\angle ADB = \angle ADC = 90^\circ$ (Since $AD \perp BC$)
  • $AD = AD$ (Common side)
• By SAS congruence criterion, $\triangle ABD \cong \triangle ACD$.
• Therefore, $AB = AC$ by CPCT.
• Since two sides of $\triangle ABC$ are equal, $\triangle ABC$ is an isosceles triangle. [Hence Proved]

Q3. 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 $\triangle ABE$ and $\triangle ACF$:
  • $\angle AEB = \angle AFC = 90^\circ$ (Altitudes)
  • $\angle A = \angle A$ (Common angle)
  • $AB = AC$ (Given)
• By AAS congruence criterion, $\triangle ABE \cong \triangle ACF$.
• Therefore, $BE = CF$ by CPCT. Thus, the altitudes are equal. [Hence Proved]

Q4. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that:
(i) ▵ABE ≅ ▵ACF
(ii) AB = AC, i.e., ABC is an isosceles triangle.

Proof:

• (i) In $\triangle ABE$ and $\triangle ACF$:
  • $\angle AEB = \angle AFC = 90^\circ$ (Given altitudes)
  • $\angle A = \angle A$ (Common angle)
  • $BE = CF$ (Given)
• By AAS congruence criterion, $\triangle ABE \cong \triangle ACF$. [Hence Proved]
• (ii) Since $\triangle ABE \cong \triangle ACF$, we have $AB = AC$ by CPCT. Thus, $ABC$ is an isosceles triangle. [Hence Proved]

Q5. ABC and DBC are two isosceles triangles on the same base BC. Show that ∠ABD = ∠ACD.

Proof:

• Let's join the vertices $A$ and $D$ to form a line segment $AD$.
• In $\triangle ABD$ and $\triangle ACD$:
  • $AB = AC$ (Given, $\triangle ABC$ is isosceles)
  • $DB = DC$ (Given, $\triangle DBC$ is isosceles)
  • $AD = AD$ (Common side)
• By SSS congruence criterion, $\triangle ABD \cong \triangle ACD$.
• Therefore, $\angle ABD = \angle ACD$ by CPCT. [Hence Proved]

Q6. ▵ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB. Show that ∠BCD is a right angle.

Proof:

• In $\triangle ABC$, since $AB = AC$, we have:
  $\angle ACB = \angle ABC$     --- (Equation 1)
• Given $AD = AB$, and since $AB = AC$, we have $AD = AC$.
• In $\triangle ACD$, since $AD = AC$, we have:
  $\angle ACD = \angle ADC$     --- (Equation 2)
• Adding Equation 1 and Equation 2:
  $\angle ACB + \angle ACD = \angle ABC + \angle ADC$
  $\implies \angle BCD = \angle ABC + \angle ADC$     --- (Equation 3)
• In $\triangle BCD$, sum of all angles is $180^\circ$:
  $\angle BCD + \angle DBC + \angle BDC = 180^\circ$
  $\implies \angle BCD + (\angle ABC + \angle ADC) = 180^\circ$ (Note that $\angle DBC = \angle ABC$ and $\angle BDC = \angle ADC$)
• Substituting Equation 3 into this equation:
  $\angle BCD + \angle BCD = 180^\circ$
  $\implies 2\angle BCD = 180^\circ$
  $\implies \angle BCD = 90^\circ$.
• Hence, $\angle BCD$ is a right angle. [Hence Proved]

Q7. ABC is a right-angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.

Solution:

• In $\triangle ABC$, $AB = AC \implies \angle C = \angle B$ (Angles opposite to equal sides).
• We know that $\angle A + \angle B + \angle C = 180^\circ$ (Angle sum property).
• Substitute $\angle A = 90^\circ$ and $\angle C = \angle B$:
  $90^\circ + \angle B + \angle B = 180^\circ$
  $\implies 2\angle B = 180^\circ - 90^\circ$
  $\implies 2\angle B = 90^\circ \implies \angle B = 45^\circ$.
• Since $\angle B = \angle C$, both angles are $45^\circ$.
• Answer: $\angle B = 45^\circ$ and $\angle C = 45^\circ$.

Q8. Show that the angles of an equilateral triangle are 60° each.

Proof:

• Let $ABC$ be an equilateral triangle. Therefore, $AB = BC = CA$.
• Since $AB = AC \implies \angle C = \angle B$ (Angles opposite to equal sides).
• Since $BC = AC \implies \angle A = \angle B$.
• From the above two relations, we get $\angle A = \angle B = \angle C$.
• By Angle Sum Property:
  $\angle A + \angle B + \angle C = 180^\circ$
  $\implies 3\angle A = 180^\circ \implies \angle A = 60^\circ$.
• Since all three angles are equal, $\angle A = \angle B = \angle C = 60^\circ$. [Hence Proved]

Aptitude Corner (For NTSE & Olympiad Aspirants)

Practice Question: In an isosceles triangle ABC with $AB = AC$, the bisectors of external angles at B and C meet at point P. If $\angle A = 80^\circ$, find $\angle BPC$.

Solution Hint: First find internal angles $\angle B$ and $\angle C$ using $180^\circ$ properties. $\angle B = \angle C = (180^\circ - 80^\circ)/2 = 50^\circ$. Exterior angles at B and C will be $180^\circ - 50^\circ = 130^\circ$ each. Their half-bisectors inside $\triangle BPC$ will be $65^\circ$ each. Therefore, $\angle BPC = 180^\circ - (65^\circ + 65^\circ) = 50^\circ$.

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