1. In triangle $A B C$, there is a point $D$ on side $B C$ such that $B A=A D=D C$. Suppose $\angle B A D=80^{\circ}$. Determine the size of $\angle A C B$.
Solution 1: Let $x=\angle A D B$. Then since $A B=A D, \angle A B D=x$. Summing the angles of $\triangle A B D$ gives $2 x+80^{\circ}=180^{\circ}$. So $2 x=100^{\circ}$ and $x=50^{\circ}$. Therefore, $\angle A D B=50^{\circ}$. Hence, $\angle A D C=180^{\circ}-\angle A D B=180^{\circ}-50^{\circ}=130^{\circ}$.
Since $A D=D C, \angle A C D=\angle D A C$. Let $y$ be this common angle. Then summing the angles of $\triangle A C D$ yield $2 y+130^{\circ}=180^{\circ}$. Therefore, $2 y=50^{\circ}$. Hence, $y=25^{\circ}$. Therefore, $\angle A C B=\angle A C D=25^{\circ}$.
Solution 2: Since $A B=A D, \angle A B D=\angle A D B$. Similarly, since $D A=D C, \angle D A C=$ $\angle D C A$. Let $x=\angle A B D=\angle A D B$ and $y=\angle D A C=\angle D C A$. Since $\angle D C A=\angle B C A$ we need to determine the value of $y$.
By external angle theorem on $\triangle A D C, x=2 y$. Summing the angles of $\triangle A B D$ gives the equation $4 y+80^{\circ}=180^{\circ}$. Therefore, $4 y=100^{\circ}$, and $y=25^{\circ}$.