Adventitious Quadrangle: Linear Case

geometry

Looks like no more answers, so I am posting my solutions.
I found a direct method using inscribed angles, and solutions to three problems are very similar.

$\mathrm{E}$: intersection between the circumscribed circle of △$\mathrm{ACD}$ and line $\mathrm{BC} \ (\ne \mathrm{C})$
$\mathrm{O}$: circumcenter of △$\mathrm{BED}$

Case i)

$(a, \ b, \ c, \ d, \ e) = (x, \ x, \ 60^\circ-x, \ x-30^\circ, \ 120^\circ-2x)$

$\angle \mathrm{CED} = 180^\circ - \angle \mathrm{CAD} = 180^\circ - x$
$\angle \mathrm{BDE} = 180^\circ - \mathrm{(\angle BED + \angle DBE)} = 30^\circ$
$\angle \mathrm{BOE} = 2\angle \mathrm{BDE} = 60^\circ \ \Rightarrow \ $ △$\mathrm{BOE} \ $ equilateral
$\angle \mathrm{BOD} = 2(180^\circ - \angle \mathrm{BED}) = 2x = \angle \mathrm{BAD} \ \Rightarrow \ \mathrm{ADBO} \ $ cyclic, $\ \angle \mathrm{OBD} = 90^\circ - x$
$\angle \mathrm{ABO} = \angle \mathrm{OBD - \angle ABD} = 30^\circ \ \Rightarrow \ \mathrm{AB} \ $ is a perpendicular bisector of $\ \mathrm{OE}$
$\angle \mathrm{AEO = \angle AOE = \angle AOD + (\angle BOD - 60^\circ) = \angle ABD} + (2x - 60^\circ) = x$
$\therefore \ \angle \mathrm{ACD = \angle AED = \angle BED - (60^\circ + \angle AEO}) = 120^\circ - 2x$

Case1

Case ii)

$(a, \ b, \ c, \ d, \ e) = (x, \ x, \ 90^\circ-3x, \ x+30^\circ, \ 60^\circ-2x)$

$\angle \mathrm{CED = \angle CAD} = x$
$\angle \mathrm{BDE} = \mathrm{\angle CBD - \angle BED} = 30^\circ$
$\angle \mathrm{BOE} = 2\angle \mathrm{BDE} = 60^\circ \ \Rightarrow \ $ △$\mathrm{BOE} \ $ equilateral
$\angle \mathrm{ABO} = 180^\circ - (\angle \mathrm{ABC} + 60^\circ) = 2x = \angle \mathrm{BAD} \ \Rightarrow \ \mathrm{AD \parallel OB}$
$\angle \mathrm{DOB = 2 \angle DEB} = 2x = \angle \mathrm{ABO}$
$\ \Rightarrow \ \mathrm{ADBO} \ $ is an isoceles trapezoid $\ \Rightarrow \ \angle \mathrm{OEA = \angle BED} = x$ $\therefore \ \angle \mathrm{ACD = \angle AED = 60^\circ - (\angle BED + \angle OEA}) = 60^\circ - 2x$

Case2

Case iii)

$(a, \ b, \ c, \ d, \ e) = (x, \ x, \ 45^\circ-1.5x, \ x+30^\circ, \ 30^\circ-x)$

$\angle \mathrm{CED} = \angle \mathrm{CAD} = x$
$\angle \mathrm{BDE} = \mathrm{\angle CBD - \angle BED} = 30^\circ$
$\angle \mathrm{BOE} = 2\angle \mathrm{BDE} = 60^\circ \ \Rightarrow \ $ △$\mathrm{BOE} \ $ equilateral
$\angle \mathrm{BOD} = 2\angle \mathrm{BED} = 2x = \angle \mathrm{BAD} \ \Rightarrow \ \mathrm{ADBO} \ $ cyclic
$\angle \mathrm{ABO} = 180^\circ - (\angle \mathrm{ABC} + 60^\circ) = 45^\circ + 0.5x$
$\angle \mathrm{AOB = \angle AOD + \angle BOD = \angle ABD} + 2x = 45^\circ + 0.5x = \angle \mathrm{ABO}$
$\ \Rightarrow \ \mathrm{AB = AO} \ \Rightarrow \ \mathrm{AE} \ $ is a perpendicular bisector of $\ \mathrm{BO}$
$\therefore \ \angle \mathrm{ACD = \angle AED = \angle BEA - \angle BED} = 30^\circ - x$