I3.1 In Figure 1 , a rectangle is sub-divided into 3 identical squares of side length 1 .
If $\alpha^{\circ}=\angle A B D+\angle A C D$, find the value of $\alpha$.
Method 1 (compound angle)
$$
\begin{array}{l}
\tan \angle A B D=\frac{1}{3}, \tan \angle A C D=\frac{1}{2} \\
0^{\circ}0 \\
\alpha=45 \\
\text { Method } 2 \text { (congruent triangles) }
\end{array}
$$

Method 2 (congruent triangles)
Draw 3 more identical squares $B C F E, C I G H, I D H G$ of length 1 as shown in the figure. $A L B D, D B E H$ are identical rectangles. Join $B G, A G$.
$B E=G H=A D$ (sides of a square)
$E G=H A=C D$ (sides of 2 squares)
$\angle B E G=90^{\circ}=\angle G H A=\angle A D C$ (angle of a square)
$\triangle B E G \cong \triangle G H A \cong \triangle A D C$ (SAS)
Let $\angle B G E=\theta=\angle G A H$ (corr. $\angle \mathrm{s} \cong \Delta$ 's)
$\angle A G H=90^{\circ}-\theta(\angle \mathrm{s}$ sum of $\triangle)$
$\angle A G B=180^{\circ}-\angle A G H-\angle B G E$ (adj. $\angle$ s on st. line)
$$
=180^{\circ}-\theta-\left(90^{\circ}-\theta\right)=90^{\circ}
$$
$B G=A G$ (corr. sides $\cong \Delta ' s$ s)
$\angle A B G=\angle B A G$ (base $\angle$ s isos. $\Delta$ )
$$
=\frac{180^{\circ}-90^{\circ}}{2}=45^{\circ}(\angle \mathrm{s} \text { sum of } \triangle)
$$
$\alpha^{\circ}=\angle A B D+\angle A C D$
$=\angle A B D+\angle G B I$ (corr. $\angle \mathrm{s}, \cong \triangle$ 's)
$$
=45^{\circ} \Rightarrow \alpha=45
$$

Method 3 (similar triangles)
Join $A I$.
$A I=\sqrt{2}$ (Pythagoras' theorem)
$\frac{B I}{A I}=\frac{2}{\sqrt{2}}=\sqrt{2}, \frac{A I}{C I}=\frac{\sqrt{2}}{1}=\sqrt{2}$
$\angle A I B=\angle C I A$ (common $\angle$ )
$\triangle A I B \sim \triangle C I A$ (2 sides proportional, included $\angle$ )
$\therefore \angle A C D=\angle B A I$ (corr. $\angle \mathrm{s}, \sim \Delta$ 's)
$\alpha^{\circ}=\angle A B D+\angle A C D=\angle A B I+\angle B A I$
$=\angle A I D($ ext. $\angle$ of $\triangle)$
$=45^{\circ}$ (diagonal of a square) $\Rightarrow \alpha=45$

Method 4 (vector dot product)
Define a rectangular system with $\overrightarrow{B C}=\mathbf{i}, \overrightarrow{B L}=\mathbf{j}$.
$\overrightarrow{A B}=-3 \mathbf{i}-\mathbf{j}, \overrightarrow{A C}=-2 \mathbf{i}-\mathbf{j}, \overrightarrow{A I}=-\mathbf{i}-\mathbf{j}$
$\overrightarrow{A B} \cdot \overrightarrow{A I}=|\overrightarrow{A B}||\overrightarrow{A I}| \cos \angle B A I$
$\cos \angle B A I=\frac{(-3)(-1)+(-1)(-1)}{\sqrt{(-3)^{2}+(-1)^{2}} \cdot \sqrt{(-1)^{2}+(-1)^{2}}}=\frac{2}{\sqrt{5}}$
$\cos \angle A C D=\frac{2}{\sqrt{5}} \Rightarrow \angle B A I=\angle A C D$
$\alpha^{\circ}=\angle A B D+\angle A C D=\angle A B I+\angle B A I=\angle A I D($ ext. $\angle$ of $\Delta) \Rightarrow \alpha=45$
Method 5 (complex number)
Define the Argand diagram with $B$ as the origin, $B D$ as the real axis, $B L$ as the imaginary axis.
Let the complex numbers represented by $A B$ and $A C$ be $z_{1}$ and $z_{2}$ respectively.
$$
\begin{array}{l}
z_{1}=3+i, z_{2}=2+i \\
z_{1} \cdot z_{2}=(3+i)(2+i)=6-1+(2+3) i=5+5 i \\
\operatorname{Arg}\left(z_{1} z_{2}\right)=\operatorname{Arg}\left(z_{1}\right)+\operatorname{Arg}\left(z_{2}\right) \\
\alpha^{\circ}=\angle A B D+\angle A C D=\tan ^{-1} \frac{5}{5}=45^{\circ} \\
\alpha=45
\end{array}
$$