Right Tangential Trapezoid with an incircle: Find the radius of the incircle and the area of the trapezoid The incircle of the trapezoid $ABCD$ that has a right angle $A$ with the center of $O$ is tangent with the point $E$ on $CD$ in a way that $CE=3$ and $ED=12$.
Find the angle of $COD$, the radius of the incircle and the area of the trapezoid.
 A: Suppose the $AD$ and $BC$ are the bases of the trapezoid $ABCD$:

According to the data given, $CE = 3$ and $DE = 12$
Suppose $\angle ADC = 2\alpha$. Since $\angle BAD = 90^\circ$, $\angle ADC + \angle DCB = 180^\circ = 2\alpha + \angle DCB$:
$$\therefore \, \angle DCB = 180^\circ - 2\alpha = 2(90^\circ - \alpha) \tag1$$
Since $AD$ and $DC$ are tangent to the inscribed circle, $DY = DE = 12$. Simillarly, considering tangents $CB$ and $CD$, $CE = CX = 3$.
Using the knowledge of geometry, we can prove that $\triangle DOY$ and $\triangle DOE$ are equivalent, therefore, $\angle ODY = \angle ODE = \frac{1}{2} \angle EDY = \alpha$. Simillarly, $\triangle COX$ and $\triangle COE$ are also equivalent, and therefore, $\angle OCX = \angle OCE = \frac{1}{2} \angle ECX = 90^\circ - \alpha$ (from the equation $(1)$)).
Now, consider the $\triangle COD$. We have proved that  $\angle ODC = \alpha$ and $\angle OCD = 90^\circ - \alpha$.
$$\therefore \, \angle COD = 180^\circ - (\alpha + 90^\circ - \alpha) = 90^\circ \tag2$$
To find the radius $(r)$ of the incircle,  we should consider right triangles, $\triangle ODE$ and $\triangle COE$. They are equal angled triangles:
$$\angle ODE = \angle COE; \quad  \angle DOE = \angle ECE; \quad \angle OED = \angle OEC$$
$$\therefore \, \frac{OY}{CE} = \frac{DY}{OE} = \frac{OD}{OC} \ \Rightarrow \ \frac{r}{3} = \frac{12}{r} = \frac{OD}{OC} \tag3$$
From the equation $(3)$,
$$\frac{r}{3} = \frac{12}{r}  \ \Rightarrow \ r^2 = 3 \times 12 = 36  \ \Rightarrow \ \therefore \ r = \sqrt{36} = 6$$
The area of the trapezoid $ABCD$:
$$\text{Area} = \frac{1}{2}(AD + BC)\cdot AB =  \frac{1}{2}(r + 12 + r + 3)\cdot 2r =  \frac{1}{2}(2r + 15)\cdot 2r = r(2r + 15) \\ = 6 \times (12 + 15) = 162$$
