Problem 1301

Solution 1: (1) As shown in Figure 12-6(1), construct $\angle CBE = \angle CAD, \angle ACD = \angle BCE$, and let sides $BE$ and $CE$ intersect at $E$. Thus, $\triangle ACD \sim \triangle BCE$, which implies
$$\frac{AC}{BC} = \frac{AD}{BE} = \frac{CD}{CE}$$

Therefore, $AC \cdot BE = BC \cdot AD = AC \cdot BD$, from which we can deduce that $BE = BD$.
Since $\angle ADB = \angle CBD + \angle CAD + \angle ACB = 90^{\circ} + \angle ACB$,
it follows that $\angle DBE = \angle CBD + \angle CBE = \angle CBD + \angle CAD = 90^{\circ}$.
Thus, $BD \perp BE$, and $\triangle DBE$ is an isosceles right triangle.
From (1), we know $\frac{AC}{BC} = \frac{CD}{CE}$, and $\angle ACD = \angle BCE$, so $\angle ACB = \angle DCE$. Therefore, $\triangle CAB \sim \triangle CDE$, and $\frac{DE}{AB} = \frac{CE}{CB} = \frac{CD}{CA}$. Consequently, $\frac{AB}{BD} \cdot \frac{CD}{CA} = \frac{AB}{BD} \cdot \frac{DE}{AB} = \frac{DE}{BD} = \sqrt{2}$.
(2) As shown in Figure 12-6(2), $CK$ and $CL$ are the tangents to the circumcircles of $\triangle ACD$ and $\triangle BCD$ at $C$, respectively. Then, $\angle LCD = \angle CBD$, $\angle KCD = \angle CAD$. Therefore, $\angle LCK = \angle LCD + \angle KCD = \angle CBD + \angle CAD = 90^{\circ}$, which means $CL \perp CK$.

Solution 2: (1) As shown in Figure 12-6(3), rotate $BC$ around $C$ by $90^{\circ}$ to $B'C$, and connect $BB'$, $AB'$, and $CD$. By the given conditions, $\angle ACB' = \angle ACB + 90^{\circ} = \angle ADB$, and $\frac{AC}{B'C} = \frac{AC}{BC} = \frac{AD}{BD}$. Therefore, $\triangle ACB' \sim \triangle ADB$, and $\frac{AC}{AD} = \frac{CB'}{BD} = \frac{AB'}{AB}$, $\angle CAB' = \angle BAD$. Thus, $\angle CAD = \angle BAB'$. From this, we can deduce that
$\triangle ADC \sim \triangle ABB'$, and $\frac{CD}{BB'} = \frac{AD}{AB}$. Therefore,
$$\begin{aligned} \frac{AB \cdot CD}{AC \cdot BD} & = \frac{AB}{BD} \cdot \frac{CD}{BB'} \cdot \frac{BB'}{AC} = \frac{AB}{BD} \cdot \frac{AD}{AB} \cdot \frac{BB'}{AC} \\ & = \frac{AD \cdot \sqrt{2} BC}{BD \cdot AC} = \sqrt{2}. \end{aligned}$$
(2) As shown in Figure 12-6(4), $l_1$ and $l_2$ are the tangents to the circumcircles of $\triangle ADC$ and $\triangle BDC$ at $C$, respectively. Then, $\angle DCE = \angle DAC$, $\angle DCF = \angle DBC$.
Since $\angle ADB = \angle ACB + 90^{\circ}$, it follows that $180^{\circ} - \angle DAB - \angle DBA = 180^{\circ} - \angle DAB - \angle DBA - \angle DAC - \angle DBC + 90^{\circ}$, which means $\angle DAC + \angle DBC = 90^{\circ}$. Therefore, $\angle DCE + \angle DCF = 90^{\circ}$, and hence $l_1 \perp l_2$.