2. $A B C$ is an isosceles triangle with $\overline{A C}=\overline{B C}$. Two points $A^{\prime}, A^{\prime \prime}$ are chosen on side $A C$ and a point $B^{\prime}$ on side $B C$ such that $\overline{A B}=\overline{B A^{\prime}}=\overline{A^{\prime} B^{\prime}}=\overline{B^{\prime} A^{\prime \prime}}=\overline{A^{\prime \prime} C}$. What is the measure of angle $A \widehat{C} B$?
(A) $18^{\circ}$
(B) $15^{\circ}$
(C) $24^{\circ}$
(D) $20^{\circ}$
(E) $30^{\circ}$
Problem 2. The correct answer is (D).

Let $\alpha=A \widehat{C} B$, then $A^{\prime \prime} \widehat{B}^{\prime} C=\alpha\left(B^{\prime} A^{\prime \prime} C\right.$ is isosceles) and thus $A^{\prime} \widehat{A^{\prime \prime}} B^{\prime}=2 \alpha$ (external angle of the triangle $B^{\prime} A^{\prime \prime} C$ ), so $A^{\prime \prime} \widehat{A^{\prime}} B^{\prime}=2 \alpha\left(A^{\prime} B^{\prime} A^{\prime \prime}\right.$ is isosceles). It follows that $A^{\prime} \widehat{B^{\prime}} B=3 \alpha$ (external angle of the triangle $\left.A^{\prime} B^{\prime} C\right)$, so $A^{\prime} \widehat{B} B^{\prime}=3 \alpha\left(B A^{\prime} B^{\prime}\right.$ is isosceles). From this, we get $A \widehat{A^{\prime}} B=4 \alpha$ (external angle of the triangle $\left.B A^{\prime} C\right)$, hence $A^{\prime} \widehat{A} B=4 \alpha\left(A B A^{\prime}\right.$ is isosceles), as well as $A \widehat{B} C=4 \alpha$. Summing the three angles of $A B C$, we conclude that $9 \alpha=180^{\circ}$, that is, $\alpha=20^{\circ}$.

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