4. The diagram below shows $\triangle A B C$, which is isoceles with $A B=A C$ and $\angle A=20^{\circ}$. The point $D$ lies on $A C$ such that $A D=B C$. The segment $B D$ is constructed as shown. Determine $\angle A B D$ in degrees.
4. Answer. 10
Solution. Let $E$ be the point inside $\triangle A B C$ such that $\triangle E B C$ is equilateral. Connect $A$ and $D$ to $E$ respectively.
It is clear that $\triangle A E B$ and $\triangle A E C$ are congruent, since $A E=A E, A B=A C$ and $B E=C E$. It implies that $\angle B A E=\angle C A E=10^{\circ}$.
Since $A D=B C=B E, \angle E B A=\angle D A B=20^{\circ}$ and $A B=B A$, we have $\triangle A B E$ and $\triangle B A D$ are congruent, implying that $\angle A B D_{4}=\angle B A E=10^{\circ}$.