
5. In triangle $A B C$, let $D$ be the midpoint of $B C$. If $\angle A D B=45^{\circ}$ and $\angle A C D=30^{\circ}$, determine $\angle B A D$.


Solution: Draw $B L$ perpendicular to $A C$ and join $L$ to $D$. (See Fig. 2.)

![](https://cdn.mathpix.com/cropped/2024_06_05_64791d6c161bbc1cf837g-3.jpg?height=442&width=656&top_left_y=359&top_left_x=775)

Fig. 2 .

Since $\angle B C L=30^{\circ}$, we get $\angle C B L=60^{\circ}$. Since $B L C$ is a right-triangle with $\angle B C L=30^{\circ}$, we have $B L=B C / 2=B D$. Thus in triangle $B L D$, we observe that $B L=B D$ and $\angle D B L=60^{\circ}$. This implies that $B L D$ is an equilateral triangle and hence $L B=L D$. Using $\angle L D B=60^{\circ}$ and $\angle A D B=45^{\circ}$, we get $\angle A D L=15^{\circ}$. But $\angle D A L=15^{\circ}$. Thus $L D=L A$. We hence have $L D=L A=L B$. This implies that $L$ is the circumcentre of the triangle $B D A$. Thus

$$
\angle B A D=\frac{1}{2} \angle B L D=\frac{1}{2} \times 60^{\circ}=30^{\circ}
$$
