Angle Chasing Will Not Help You Here

By angle chasing we find $\angle ECD = 30°, \angle AEC = 39°, \angle EAD = 21°, \angle CDA = \angle ACD = 48°$ . So $\triangle ACD$ is isosceles.

Let $M$ be the midpoint of $CD$ . So, $AM$ is the perpendicular bissector.

WLOG, let $AC=AD=1$ . So, $CM=MD=\cos { 48° }$ and $CD=2\cos { 48° }$ . Let $\angle DEA$ be $\alpha$ .

By sine rule in $\triangle CDE$ the get: $\frac { \sin { (39°+\alpha ) } }{ CD } =\frac { \sin { 30° } }{ ED } \Rightarrow \frac { \sin { (39°+\alpha ) } }{ 2\cos { 48° } } =\frac { \sin { 30° } }{ ED }$ .

By sine rule in $\triangle ADE$ the get: $\frac { \sin { (\alpha ) } }{ AD } =\frac { \sin { 21° } }{ ED } \Rightarrow \frac { \sin { (\alpha ) } }{ 1 } =\frac { \sin { 21° } }{ ED }$ .

Doing the math with the 2 above equations, we get: $\frac { \sin { (39°+\alpha ) } }{ \cos { 48° } } =\frac { \sin { (\alpha ) } }{ \sin { 21° } }$

Since $\cos { 48° } =\sin { 42° } =2\sin { 21° } \cos { 21° }$ we can rewrite the last equation as: $\frac { \sin { (39°+\alpha ) } }{ \sin { (\alpha ) } } =2\cos { 21° }$

Since $\sin { (39°+\alpha ) } = \sin { 39° } \cos { \alpha } +\sin { \alpha } \cos { 39° }$ we will get in left side of equation:

$\frac { \sin { 39° } \cos { \alpha } +\sin { \alpha } \cos { 39° } }{ \sin { \alpha } } =\sin { 39° } \cot { \alpha } +\cos { 39° }$

Now, we will need some creativity in right side: Write $21°$ as $(60° - 39°)$ :

$2\cos { 21° } = 2\cos { (60° - 39°) } = 2 \left( \cos { 60° } \cos { 39° } +\sin { 60° } \sin { 39° } \right) =\cos { 39° } +\sqrt { 3 } \sin { 39° }$ .

Now, the equation is:

$\sin { 39° } \cot { \alpha } +\cos { 39° } = \cos { 39° } +\sqrt { 3 } \sin { 39° }$

Then:

$\sin { 39° } \cot { \alpha } = \sqrt { 3 } \sin { 39° }$

and finally: $\cot { \alpha } = \sqrt { 3 }$ . Therefore, $\alpha = 30°$ .

Returning to the $\triangle CDE$ and plugging in $\alpha = 30°$ , we have $39° + 30° + \angle CDE + 30° = 180° \Rightarrow \angle CDE = \boxed{81°}$ .

And we are done.

A Synthetic Solution along the lines of Langley's Adventitious Angles: Draw a line AG at 24° to AC meeting BC in G. Join G to D & observe that / AGC = 180 -78-24 =78°. So in tr. AGC we've AG = AC. Now / ADC = 180 -84 -48 = 48°. Thus, AD = AC = AG, But / DAG = 84 -24=60°. Thus, triangle ADG is equilateral. Now observe that / EAG =60-21=39° while / EGA=180 -78=102°; hence / GEA=180 -102 -39 =39°. This makes triangle EGA isosceles with GE = AG = GD. Now / EGD =102 -60 =42° which, in turn, means that / GDE=/ GED= (180-42)/2 = 69° and therefore / CDE = 180 -- 69 -30 = 81°. (Incidentally, /DEA = 30°, same as in Langley)

The problem can be generalized, here are the sufficient information I will use from this configuration: $CA=DA, \angle BCD=30^{\circ}, 2\angle CEA=\angle ACB$ .

Construct $F\ne C$ on $BC$ such that $AF=AC$ , then $\angle FAE=\angle AFE-\angle FEA=\angle ACE-\angle CEA=\angle CEA$ , which means $FA=FE$ . In addition, observe that $A$ is the circumcenter of $\triangle CFD$ ; therefore $\angle FAD=2\angle DCF=60^{\circ}$ and $\triangle AFD$ is equilateral. Hence $FE=FA=FD$ , deducing that $F$ is the circumcenter of $\triangle ADE$ . Thus :

$\angle EDC=\angle EDA-\angle CDA=90^{\circ}+\angle CEA-\angle CDA$

For this special configuration, $\angle EDC=90+39-48=\boxed {81^{\circ}}$

Using Sine Rule,

$\dfrac{\sin 39^\circ\sin 84^\circ}{\sin 63^\circ\sin 48^\circ}\sin x^\circ = \sin(30^\circ+x^\circ)$

Solving, $\sin x = 0.987688\\\Large\boxed {x = 81^\circ}$

How I got it? Here:

Let $CD$ intersect $EA$ at $F$ and $EC = m\\CD = n\\CA = b\\\angle CDE = x\\\angle DEC = y$ Chasing angles, $\angle CEA = 39^\circ\\\angle ECD = 30^\circ\\ \angle CAE = 63^\circ\\\angle CDA = ~48^\circ$

Sine rule says, in a triangle, $\dfrac {a\sin B}{\sin A} = b$

So, in $\triangle CEA$ , $\dfrac {m\sin 39^\circ}{\sin 63^\circ} = b$

Now, in $\triangle CDA$ , $\dfrac{b \sin 84^\circ}{\sin48^\circ} = n\\\\\Rightarrow \dfrac {m\sin 39^\circ}{\sin 63^\circ}\cdot\dfrac{\sin 84^\circ}{\sin48^\circ}=~n$

In $\triangle EDC$ $180^\circ - (x + 30^\circ) = y\\\dfrac{m\sin y}{\sin x} = ~n$

$\sin(180^\circ - \theta) = \sin\theta$

So, we get $m\dfrac {\sin 39^\circ\sin 84^\circ}{\sin 63^\circ\sin48^\circ} = m\dfrac{\sin y}{\sin x}\\\\\Rightarrow \dfrac{\sin 39^\circ\sin 84^\circ}{\sin 63^\circ\sin 48^\circ}\sin x = \sin(x+30^\circ)$

Solve it to get the answer!

In my solution I assumed the length of AC to be 1, then I used sine law and cosine law to find for the required angle.

From what is given, <CDA = 48, so triangle ACD is isosceles. Let CA = AD = 1. By the Law of sines, CD/ sin(84) = 1/ sin(48), so CD = 1.338261212.Filling some angles, < COA = 69, where O is the intersection of EA and CD., <EAD = 21, <BCD = 30, < CEA = 39. Define < AED by t. . Then by Sine Law, ED / sin(21, so ED, = .35836795/ sin(t). Also ED/sin(30) = CD/sin(39 + t) = 1/ sin(t). Using the value for CD and equating these two equations for ED , we can solve for t; t = 30, and from triangle EDO, we have x = 180 -t - 69 = 81. Ed Gray

Let AE and CD intersect at F. Following are the angles we get.
ACD.. =.. 48,......... ADC.. =.. 48,.......... ECD.. =.. 30,.......... EAC.. =.. 63,.......... EAD.. =... 21,..... AEC.. =.. 39 .......
EFC.. =.. DFA.. =.. 111, .......... DFE.. =.. AFC.. =.. 69,.......... CDE.. =.. X ........... DEA.. =.. Y.. =.. 111.. -.. X
Let AC = AD = a, since isosceles triangle CAD has two 48 angles.
Applying sin rule to the following triangles,
ACE :- EC/sin63 = a/sin39
CED :- EC/sinX = ED/sin30 ...............> sinX/EC = sin30/ED
AED :- a/sinY = ED/sin21...............................................................................................................................

( a/sinY ) ( sinX/EC ) ( EC/sin63 ) = ( a/sin39 ) ( sin30/ED) ( ED/sin21)
sinX/{ sin( 111 - X ) } = ( sin63/ sin39 ) * ( sin30/ sin21) .........................................................................

EXPANDING sin( 111 - X ) AND SIMPLIFYING WE GET X = 81.
Sorry I am not able to copy the sketch I have with me.