Stuck on a Geometry Problem (II)

Question

(Looking for a geometric solution.)

$ABCD$ is quadrilateral, where $|AB|=|AD|=|DC|$. $AED$ is a triangle, where $E$ is the midpoint of $BC$. $\angle DAB=84 ^{\circ}$ and $\angle CDA=72 ^{\circ}$. Determine the measure of $\angle AED$.

Answer 1

Using algebra in the complex plane we place A at (0,0), D at (1,0), B at (cos(84),sin(85)), C at (cos(180-72),sin(180-72)) relative to D, where the common length of the 3 sides is set to 1. Then we compute the position of E as the mean value of B and C, the direction from E to A by subtracting A and E and the direction from E to D by subtracting D from E. Then the angle phi is given by the argument (angle) obtained by dividing ED through AD. In Maple the computation reads:

Ap := 0+I*0 ; Dp := 1+I*0 ; Bp := cos(84*Pi/180)+I*sin(84*Pi/180) ; Cp := Dp+cos((180-72)Pi/180)+Isin((180-72)*Pi/180) ; Ep := (Bp+Cp)/2 ; EAp := Ap-Ep ; EDp := Dp-Ep ; EAp/EDp/abs(EAp)*abs(EDp); evalf(%) ; argument(%)*180/Pi ; evalf(%) ;

The result is 54 degrees.