3. The triangle $A B C$ is isosceles with $A B=B C$. The point $D$ is a point on $B C$, between $B$ and $C$, so that $A C=A D=B D$.
What is the size of angle $A B C$ ?
SolUtion
Commentary
For this solution, we use the results that
(a) 'the angles opposite the two equal sides of an isosceles triangle are equal' (base angles of an isosceles triangle),
(b) 'an exterior angle of a triangle is equal to the sum of the two opposite interior angles' (exterior angle of a triangle),
(c) 'the angle sum of a triangle is $180^{\circ}$ ' (angle sum of a triangle).

Let angle $A B C$ be $x^{\circ}$, as shown.
Triangle $A B D$ is isosceles since $A D=B D$, so that $\angle A B D=\angle D A B=x^{\circ}$ (base angles of an isosceles triangle).
Thus, from triangle $A B D$, we have $\angle A D C=2 x^{\circ}$ (exterior angle of a triangle).
Triangle $A C D$ is isosceles with $A C=A D$ so that $\angle A D C=\angle A C D=2 x^{\circ}$ (base angles of an isosceles triangle).
Also, triangle $A B C$ is isosceles with $A B=B C$ so that $\angle B C A=\angle C A B=2 x^{\circ}$ (base angles of an isosceles triangle).
Now, considering triangle $A B C$, we have $2 x+x+2 x=180$ (angle sum of a triangle).
Hence $x=36$ and so angle $A B C$ is $36^{\circ}$.