Given a circle $O$ with chord $AB = 5$. Point $C$ is on the extension of $AB$ such that $BC = 4$. A tangent line from $C$ to the circle intersects the circle at point $D$. Point $D$ is on the same side of line $AB$ as the center of the circle. We need to find various lengths and ratios related to this configuration.

1. Problem Description

Given a circle $O$ with chord $AB = 5$. Point $C$ is on the extension of $AB$ such that $BC = 4$. A tangent line from $C$ to the circle intersects the circle at point $D$. Point $D$ is on the same side of line $AB$ as the center of the circle. We need to find various lengths and ratios related to this configuration.

2. Solution Steps

(1) Find $CD$.

By the tangent-secant theorem, $CD^2 = CB \cdot CA = CB \cdot (CB + BA) = 4 \cdot (4 + 5) = 4 \cdot 9 = 36$. Therefore, $CD = \sqrt{36} = 6$.

(2) Find the similar triangle to $\triangle CAD$.

$\angle CAD = \angle CDB$ (angle between tangent and chord).

Also, $\angle ACD = \angle DCB$ is given.

Therefore, $\triangle CAD \sim \triangle CDB$ (AA similarity).

Find $\frac{AD}{BD}$.

Since $\triangle CAD \sim \triangle CDB$, we have $\frac{AD}{CD} = \frac{CD}{CB} = \frac{CA}{CD}$.

Also, $\frac{AD}{BD} = \frac{CA}{CD} = \frac{CD}{CB}$. Therefore, $\frac{AD}{BD} = \frac{6}{4} = \frac{3}{2}$.

(3) Let $E$ and $F$ be the intersection points of the angle bisector of $\angle ACD$ with $BD$ and $AD$, respectively.

Find $\frac{DE}{BE}$ and $\frac{DF}{AF}$.

By the angle bisector theorem in $\triangle ACD$, $\frac{AF}{DF} = \frac{AC}{CD} = \frac{9}{6} = \frac{3}{2}$. So $\frac{DF}{AF} = \frac{2}{3}$.

By the angle bisector theorem in $\triangle BCD$, $\frac{DE}{BE} = \frac{CD}{BC} = \frac{6}{4} = \frac{3}{2}$.

Let $G$ be the intersection of $AE$ and $CD$.

We want to find $\frac{CG}{GD}$.

In $\triangle ADE$, $AF$ is the angle bisector of $\angle DAC$, and $AE$ intersects $CD$ at $G$.

However, directly finding CG/GD is difficult.

Consider $\triangle ADC$. $E$ is the intersection of the angle bisector of $\angle ACD$ with $BD$, and $F$ is the intersection with $AD$. $AE$ and $CD$ intersect at $G$.

Since $\frac{AD}{BD} = \frac{3}{2}$, $AD = \frac{3}{2}BD$.

Since $\frac{DE}{BE} = \frac{3}{2}$, $DE = \frac{3}{2}BE$. Thus, $BD = BE + DE = BE + \frac{3}{2}BE = \frac{5}{2}BE$, so $BE = \frac{2}{5}BD$.

$DE = BD - BE = BD - \frac{2}{5}BD = \frac{3}{5}BD$.

So $\frac{DE}{BD} = \frac{3}{5}$ and $\frac{BE}{BD} = \frac{2}{5}$.

Since $\triangle CAD \sim \triangle CDB$, $\angle CAD = \angle CDB$.

Also, $\angle ACD = \angle DCB$.

CE is the angle bisector of $\angle ACD$, so $\angle ACE = \angle DCE$.

Thus $\angle ACE = \angle DCE$. Since $\triangle CAD \sim \triangle CDB$, we have $CD/CB = AC/CD$, which implies $CD^2 = AC \times CB$, i.e., $6^2 = 9 \times 4=36$, ok.

$\frac{CG}{GD}$ requires more advanced knowledge or techniques. Let's try using Menelaus' theorem on $\triangle ADF$ and line $CE$.

$\frac{AE}{EF} \cdot \frac{FB}{BD} \cdot \frac{DC}{CA} = 1$. I can't proceed from here.

Let's use Ceva's theorem on triangle $\triangle ACD$ with cevians $AE$, $DF$, $CB$, $G$ inside it.

However, Ceva does not lead directly to CG/GD

Area of quadrilateral $ABEF$ as a multiple of the area of $\triangle ACD$.

Back to CG/GD. We know AC/CD=9/6=3/

2. AD/BD = 3/

2. Since $\frac{AF}{FD} = 3/2$, we get $AF = \frac{3}{5} AD$.

3. Final Answer

(1) $CD = 6$

(2) $\angle CAD = \angle CDB$, $\triangle CAD \sim \triangle CDB$, $\frac{AD}{BD} = \frac{3}{2}$

(3) $\frac{DE}{BE} = \frac{3}{2}$, $\frac{DF}{AF} = \frac{2}{3}$

The problem is missing the diagrams that would provide values for CG/GD and the ratio of areas. Given the ratios calculated, and without extra time, it's not feasible to determine exact answers for those values.