5. Let $D$ be a point on side $\overline{B C}$ of triangle $\triangle A B C$, such that $\overline{D C}=2 \overline{B D}$. Determine the angles of the triangle if $\measuredangle A B C=45^{\circ}$ and $\measuredangle A D C=60^{\circ}$.
Solution. Given $\overline{D C}=2 \overline{B D}$, $\measuredangle A B C=45^{\circ}, \measuredangle A D C=60^{\circ}$. Let $E$ be a point on $\overline{A D}$ such that $\overline{C E} \perp \overline{A D}$. Clearly, $\measuredangle E C D=90^{\circ}-60^{\circ}=30^{\circ}$, so we get $\overline{D C}=2 \overline{D E}$, i.e., $2 \overline{D E}=2 \overline{B D}$, from which $\overline{D E}=\overline{B D}$.

Furthermore,

![](https://cdn.mathpix.com/cropped/2024_06_05_edcf2076eef478aa4fe7g-03.jpg?height=384&width=524&top_left_y=707&top_left_x=959)

$$
\measuredangle B D E=180^{\circ}-\measuredangle C D E=180^{\circ}-60^{\circ}=120^{\circ}
$$

and since $\overline{D E}=\overline{B D}$, we get $\measuredangle D B E=\measuredangle D E B=30^{\circ}$.

Thus,

$$
\measuredangle E B C=\measuredangle E C B=30^{\circ} \text { i.e., } \overline{E C}=\overline{E B}
$$

Now

$$
\measuredangle E B A=\measuredangle C B A-\measuredangle E B C=45^{\circ}-30^{\circ}=15^{\circ} \text { and } \measuredangle B E A=180^{\circ}-30^{\circ}=150^{\circ}
$$

so it follows that $\measuredangle E A B=15^{\circ}$, i.e., $\triangle E A B$ is isosceles, i.e.,

$$
\overline{B E}=\overline{A E}
$$

From (1) and (2), it follows that $\overline{A E}=\overline{E C}$, i.e., $\measuredangle E C A=\measuredangle E A C=45^{\circ}$.

Now it is easy to see that $\measuredangle B A C=60^{\circ}, \measuredangle C B A=45^{\circ}, \measuredangle B C A=75^{\circ}$.

## VII Section