G2.4 In Figure $1, P, B$ and $C$ are points on a circle with centre $O$ and diameter $B C$. If $A, B, C$, $D$ are collinear such that $A B=B C=C D$, $\alpha=\angle A P B$ and $\beta=\angle C P D$, find the value of $(\tan \alpha)(\tan \beta)$
$$
\begin{array}{l}
\text { Let } A B=x=B C=C D, \angle C B P=\theta \text {. } \\
\left.\angle B P C=90^{\circ} \text { ( } \angle \text { in semi circle }\right), \angle B C P=90^{\circ}-\theta(\angle \mathrm{s} \text { sum of } \triangle) \\
B P=x \cos \theta, C P=x \sin \theta \\
\angle B A P=\theta-\alpha, \angle C D P=90^{\circ}-\theta-\beta \text { (ext. } \angle \text { of } \triangle \text { ) } \\
\frac{x}{\sin \alpha}=\frac{B P}{\sin \angle B A P} \text { (sine rule on } \triangle A B P \text { ); } \frac{x}{\sin \beta}=\frac{C P}{\sin \angle C D P} \text { (sine rule on } \triangle C D P \text { ) } \\
\frac{x}{\sin \alpha}=\frac{x \cos \theta}{\sin (\theta-\alpha)} ; \frac{x}{\sin \beta}=\frac{x \sin \theta}{\cos (\theta+\beta)} \\
\sin \theta \cos \alpha-\cos \theta \sin \alpha=\cos \theta \sin \alpha ; \cos \theta \cos \beta-\sin \theta \sin \beta=\sin \theta \sin \beta \\
\sin \theta \cos \alpha=2 \cos \theta \sin \alpha ; \cos \theta \cos \beta=2 \sin \theta \sin \beta \\
\tan \alpha=\frac{\tan \theta}{2} ; \tan \beta=\frac{1}{2 \tan \theta} \\
(\tan \alpha)(\tan \beta)=\frac{\tan \theta}{2} \cdot \frac{1}{2 \tan \theta}=\frac{1}{4} \\
\text { Method } 2 \angle B P C=90^{\circ} \text { ( } \angle \text { in semi circle), } \\
\text { Produce } P B \text { to } E \text { so that } P B=B E \text {. } \\
\text { Produce } P C \text { to } F \text { so that } P C=C F \text {. } \\
\because A B=B C=C D \text { (given) } \\
\therefore A P C E, B P D F \text { are //-grams (diagonals } \\
\text { bisect each other) } \\
\angle P E C=\alpha \text { (alt. } \angle \mathrm{s}, A P / / E C \text { ) } \\
\angle P F B=\beta \text { (alt. } \angle \mathrm{s}, P D / / B F \text { ) } \\
\text { In } \triangle E P C, \tan \alpha=\frac{P C}{P E}=\frac{P C}{2 P B} \\
\text { In } \triangle B P F, \tan \beta=\frac{P B}{P F}=\frac{P B}{2 P C} \\
(\tan \alpha)(\tan \beta)=\frac{P C}{2 P B} \cdot \frac{P B}{2 P C}=\frac{1}{4} \\
\end{array}
$$

Method 3 Lemma A Given a triangle $A B C$. $D$ is a point on $B C$ such that $B D: D C=m: n, A D=t$.
$$
\angle A B D=\alpha, \angle A D C=\theta<90^{\circ}, \angle A C D=\beta \text {. }
$$

Then $n \cot \alpha-m \cot \beta=(m+n) \cot \theta$
Proof: Let $H$ be the projection of $A$ on $B C$.
$A H=h, D H=h \cot \theta$.
$B H=\cot \alpha, C H=h \cot \beta$
$$
\begin{array}{l}
\frac{m}{n}=\frac{B D}{D C}=\frac{B H-D H}{C H+H D}=\frac{h \cot \alpha-h \cot \theta}{h \cot \beta+h \cot \theta} \\
m(\cot \beta+\cot \theta)=n(\cot \alpha-\cot \theta) \\
\therefore n \cot \alpha-m \cot \beta=(m+n) \cot \theta
\end{array}
$$

Lemma B Given a triangle $A B C . D$ is a point on $B C$ such that $B D: D C=m: n, A D=t$.
$$
\angle B A D=\alpha, \angle A D C=\theta<90^{\circ}, \angle C A D=\beta \text {. }
$$

Then $m \cot \alpha-n \cot \beta=(m+n) \cot \theta$
Proof: Draw the circumscribed circle $A B C$.
Produce $A D$ to cut the circle again at $E$.
$$
\angle B C E=\alpha, \angle C B E=\beta(\angle \text { s in the same seg. })
$$
$\angle B D E=\theta<90^{\circ}$ (vert. opp. $\angle$ s)
Apply Lemma A on $\triangle B E C$.
$\therefore m \cot \alpha-n \cot \beta=(m+n) \cot \theta$
Now return to our original problem
$\angle B P C=90^{\circ}(\angle$ in semi circle $)$
Apply Lemma B to $\triangle A P C$ :
$x \cot \alpha-x \cot 90^{\circ}=(x+x) \cot \theta$
$\cot \alpha=2 \cot \theta \ldots \ldots$ (1)
Apply Lemma B to $\triangle B P C, \angle B P C=90^{\circ}-\theta$
$x \cot \beta-x \cot 90^{\circ}=(x+x) \cot \left(90^{\circ}-\theta\right)$
$\cot \beta=2 \tan \theta \ldots \ldots$ (2)
(1) $\times(2): \cot \alpha \cot \beta=2 \cot \theta \times 2 \tan \theta=4$
$$
\therefore(\tan \alpha)(\tan \beta)=\frac{1}{4}
$$

Remark: the original question 圓有直徑 $B C$, 圓心在 $O, P 、 B$ 及 $C$ 皆為圓周上的點。若 $A B=B C=C D$ 及 $A D$ 為一絨段 $\ldots A B=B C=C D$ and $A D$ is a line segment $\ldots$
Both versions are not smooth and clear. The new version is as follow:
$B C$ 是圓的直徑, 圓心為 $O, P 、 B$ 及 $C$ 皆為圓周上的點。若 $A 、 B 、 C$ 及 $D$ 共絨且 $A B=$ $B C=C D \ldots$ If $A, B, C, D$ are collinear such that $A B=B C=C D \ldots$