Isosceles Triangles

Theorem 1 (Pons Asinorum):

Statement: In an isosceles triangle, the angles opposite to the equal sides are equal.

Proof:

Let $\triangle ABC$ have $AB = AC$. We must prove that $\angle B = \angle C$.

Construction: Draw the bisector of $\angle A$, meeting base $BC$ at point $D$.

In $\triangle ABD$ and $\triangle ACD$:

1. $AB = AC$ (Given).
2. $\angle BAD = \angle CAD$ (By construction, since $AD$ bisects $\angle A$).
3. $AD = AD$ (Common side).

By SAS Congruence Criterion: $\triangle ABD \cong \triangle ACD$.

By CPCTC: $\mathbf{\angle B = \angle C}$. $\blacksquare$

Theorem 2 (Converse of Pons Asinorum):

Statement: If two angles of a triangle are equal, then the sides opposite to them are also equal.

Proof:

Let $\triangle ABC$ have $\angle B = \angle C$. Draw $AD \perp BC$.

In $\triangle ADB$ and $\triangle ADC$:

1. $\angle B = \angle C$ (Given).
2. $\angle ADB = \angle ADC = 90^\circ$ (By construction).
3. $AD = AD$ (Common side).

By AAS Congruence Criterion: $\triangle ADB \cong \triangle ADC$.

By CPCTC: $\mathbf{AB = AC}$. $\blacksquare$

In $\triangle ABC$ with $AB = AC$, let line segment $AD$ be drawn from the apex vertex $A$ to base $BC$. Because $\triangle ABD \cong \triangle ACD$, the following four geometric lines are strictly identical:

• $AD$ is the altitude to $BC$ ($AD \perp BC$, since $\angle ADB = \angle ADC = 90^\circ$).
• $AD$ is the median to $BC$ ($BD = DC$).
• $AD$ is the internal bisector of the vertical angle $A$ ($\angle BAD = \angle CAD$).
• $AD$ is the perpendicular bisector of the base $BC$.

An equilateral triangle is a regular polygon with three equal sides: $AB = BC = CA$.

• Since $AB = AC \implies \angle B = \angle C$.
• Since $AB = BC \implies \angle C = \angle A$.
• Therefore, $\angle A = \angle B = \angle C$.
• By the Angle Sum Property: $\angle A + \angle B + \angle C = 180^\circ \implies 3\angle A = 180^\circ \implies \mathbf{\angle A = \angle B = \angle C = 60^\circ}$.
• Every equilateral triangle is equiangular, and every equiangular triangle is equilateral.

Problem 1: In $\triangle ABC$, the bisectors of $\angle B$ and $\angle C$ meet at $O$. If $AB = AC$, prove that: (i) $OB = OC$, (ii) $AO$ bisects $\angle A$.

Solution:

(i) Prove $OB = OC$:

Since $AB = AC$, by Pons Asinorum: $\angle B = \angle C$.

Dividing by $2$: $\frac{1}{2}\angle B = \frac{1}{2}\angle C$.

Since $OB$ and $OC$ are angle bisectors: $\angle OBC = \angle OCB$.

In $\triangle OBC$, since angles opposite to sides are equal, by the converse theorem: $\mathbf{OB = OC}$. [Proved (i)]

(ii) Prove $AO$ bisects $\angle A$:

In $\triangle ABO$ and $\triangle ACO$:

1. $AB = AC$ (Given).
2. $OB = OC$ (Proved above).
3. $AO = AO$ (Common side).

By SSS Congruence Criterion: $\triangle ABO \cong \triangle ACO$.

By CPCTC: $\angle BAO = \angle CAO$.

Hence, $AO$ bisects $\angle A$. $\blacksquare$ [Proved (ii)]