Angle bisector

Let the point where $AC$ and $BD$ intersect be $O$ . Also let $\angle BAE = \angle EAC = x$ and $\angle BDE = \angle EDC = y$ .

Since $\angle AOB = \angle DOC$ , we can write $2y+42=2x+26$ and then $\ldots y=x-8$ .

We can also write $\angle AED +x+y+(180-(2x+26))=180$ . This simplifies to $\angle AED = 34^\circ$ .

Let O be the intersection of AC and BD.

Since $\angle AOB$ and $\angle COD$ are vertical angles, then

$\angle ABD$ + $\angle BAC$ = $\angle ACD$ + $\angle BDC$ .

Hence $\angle BAC$ = 42 and $\angle BDC$ = 26. This makes $\angle EAC$ = 26 and $\angle EDB$ = 13.

$\angle AOB$ is an exterior angle of $\triangle AOD$ ,

$\angle AOB$ = $\angle OAD$ + $\angle ODA$ = 112.

$\angle AED$ = (180 - ( $\angle EAC$ + $\angle EDB$ + $\angle OAD$ + $\angle ODA$ )

$\angle AED$ = 34.

Let AC and BD intersect at P, and let ∠APB = x, we get ∠EAP = (180 - 26 - x)/2 and ∠EDP = (180 - 42 - x)/2 and reflex angle ∠APD = 180 + x. Because APDE is also a quadrilateral, hence we get ∠AED = 33.

By property the angle bisector is the midpoint of variation of angles subtended by AD on BC......there fore angle AED =(42+26)/2=34

One low-tech solution is to use a rectangle as your convex quadrilateral.

Let $F = AE \cap BD$ , $G = BD \cap AC$ , then $$\angle AGB = \angle CGD \tag{1}$$ but $$\angle AGB = 180^{\circ} - 2 \angle EAB - 26^{\circ}$$ $$\angle CGD = 180^{\circ} - 2 \angle EDC - 42^{\circ}$$ Hence in equation $(1)$ we get $$180^{\circ} - 2 \angle EAB - 26^{\circ} = 180^{\circ} - 2 \angle EDC - 42^{\circ}$$ which yields $$\angle EDC = \angle EAB - 8^{\circ}$$ Now $$\angle DFE = \angle AFB = 180^{\circ} - \angle FAB - 26^{\circ} = 154^{\circ} - \angle EAB$$ Hence $$\angle AED = \angle FED = 180^{\circ} - \angle DFE - \angle FDE$$ $$\angle AED = 180^{\circ} - (154^{\circ} - \angle EAB) - \angle EDC$$ $$\angle AED = 26^{\circ} + \angle EAB - \angle EDC$$ $$\angle AED = 26^{\circ} + 8^{\circ} = \boxed{34^{\circ}}$$