Geometry problem

Note that $\angle BAD = \angle CAD = \dfrac12 \cdot 30^\circ = 15^\circ$.

Considering the interior angle sum of $\triangle ADC$ gives $\angle ADC = 120^\circ$.

Using Sine Law in $\triangle ADC$ gives 

$\dfrac{10}{\sin 120^\circ} = \dfrac{AD}{\sin 45^\circ}\\ AD = \dfrac{10 \sqrt 6}3$

Considering the angles around point D gives $\angle ADB = 60^\circ$. Considering the interior angle sum of $\triangle ABD$ gives $\angle ABD = 105^\circ$.

Using Sine Law in $\triangle ABC$ gives 

$\dfrac{10}{\sin 105^\circ} = \dfrac{AB}{\sin 45^\circ}\\ AB = 10(\sqrt 3 - 1)$

Then, the area of $\triangle ABD$ is $\dfrac12 (AB)(AD) \sin 15^\circ = \dfrac12 (10(\sqrt 3 - 1))\cdot \dfrac{10 \sqrt 6}3 \cdot \dfrac{\sqrt 6 - \sqrt 2}4 = \dfrac{50(2 \sqrt 3 - 3)}{3}$.

Angle  ABC  =105

Let AD = x    CD = 10 - x

BC / sin 30  = AC /sin 105

BC /sin30 = 10 /sin105

BC = 5/sin105

BA / sin 45 =  AC / sin 105

BA / sin45 = 10 /sin105

BA = (10/sqrt 2) / sin 105  =   20 / (1 + sqrt 3)

Since BD is a  bisector, then

BA / AD = BC / CD

(10/sqrt2)/ sin 105 / x  = 5/sin105 / (10-x)

(10/sqrt 2) (10-x) / sin 105  =  5x / sin105

(10/sqrt 2) (10 -x)  = 5x

100/sqrt 2  - (10/sqrt 2)x = 5x

100/sqrt 2  = (5 + 10/sqrt 2) x

x = (100/sqrt 2) / ( 5 + 10/sqrt 2)  =   (100) / (5sqrt 2 + 10)  =  20 / (2 + sqrt 2) = AD

[ABD ] = (1/2)BA * AD  * sin 30  =  (1/4) [20 / (1 + sqrt 3)] * [ 20 /(2 + sqrt 2) ]  ≈ 10.72