TITLE: Find the missing angle in triangle
QUESTION [14 upvotes]: In the below triangle, we are looking for the value of angle $φ$.
We are given $α=30, β=18, γ=24$ and also that $CD=BD$.
I have solved it with trigonometry (sine law) and found the required angle to be 78 but I need to solve it with Geometry only.

What I have tried so far:
First of all, the angle is constructible, which means to me that there must be a geometrical solution. I first drew triangle ABC; easy, since we know 2 of its angles. We are not interested in the lengths of the sides. Then, with side AC as a base, and angle of 24 degrees, we can draw a ray from the point A.
Then, since $CD=BD$, triangle DCB is isosceles, therefore D must lie on the perpendicular bisector of CB, which we can draw. The point of intersection of the ray from A and the perpendicular bisector, is point D.
From triangle FEB we have that
angle AFD = 108.
From triangle AFD,
$ADC+CDE+54+108=180$ so $ADC+CDE=18$
We also have $24+ACD+ADC=180$
$ACB=132$
$132+φ+ACD=180$
$18+φ+54+ADC+2CDE=180$
I am always one equation short.
Any ideas?
Many thanks in anticipation!
EDIT:
Sine law in triangle ABD:
$\frac {sin (φ+18)}{AD} = \frac {sin (54)}{BD}$
Sine law in triangle ACD:
$\frac {sin (360-132-φ)}{AD} = \frac {sin (24)}{CD} = \frac {sin (24)}{BD}$
so
$\frac {sin (φ+18)}{sin (228-φ)} = \frac {sin (54)}{sin (24)}$
hence $φ=78$.

REPLY [4 votes]: Since $\angle DAB=54^o$, if we construct a regular pentagon on $AD$, then $AB$ bisects $\angle DAG=108^o$, and $AB$ extended to $K$ on the circumcircle passes through center $N$.
Extend $AC$ to $I$, $DB$ to $L$, and join $IK$, $KL$, $LA$, $IL$, and $DG$.
Since cyclic quadrilateral $AIKL$ has a right angle at $I$, it is a rectangle.
Therefore $\angle AIL=\angle IAK=30^o$, $\angle LAK=60^o$, and$$\angle LAG=\angle LAK-\angle GAK=60^o-54^o=6^o=\angle LDG$$And since in the regular pentagon $\angle ADG=36^o$, and as OP notes $\angle ADE=18^o$, then $\angle LDG=\angle ADC$.

Therefore$$\angle CDB=\angle ADG-2\angle LDG=36^o-2\cdot6^o=24^o$$and$$\angle DCB=\phi=\frac{180^o-24^o}{2}=78^o$$