3. Let $D$ be a point on side $B C$ of triangle $A B C$, such that $D C=2 B D$. Determine the remaining angles of triangle $A B C$ if $\measuredangle A B C=45^{\circ}$ and $\measuredangle A D C=60^{\circ}$.
Solution. Let $E$ be the foot of the perpendicular dropped from $C$ to the line $A D$ (diagram on the right). Then $2 E D = D C$, so from $2 B D = D C$ it follows that $E D = B D$. In the isosceles triangle $E B D$, the angle at the vertex is equal to $120^{\circ}$, so

$$
\measuredangle E B D = \measuredangle D E B = 30^{\circ}, \angle A B E = 15^{\circ}
$$

In triangle $E B C$, the angles at vertices $B$ and $C$ are both $30^{\circ}$, so $E B = C E$. In triangle $A B E$, the angles at vertices $A$ and $B$ are both $15^{\circ}$, so $B E = E A$. Therefore, $C E = E A$, which means triangle $C E A$ is isosceles and, being a right triangle, its acute angles are each $45^{\circ}$. From this, it follows that $\measuredangle B C A = 75^{\circ}$ and $\measuredangle C A B = 60^{\circ}$.

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