Find unknown angles of a triangle!!!

Question

I am unable to find out the unknown angles for the following triangle which I attached with this post.

Angle BAD and angle BCD are the unknown angles need to be calculated. Given that lines AB=BC=CD and angle CDE = 108 degrees

From my calculations: angle ADC = 180 - 108 = 72 degrees (angles on a straight line)

angle BAD + angle BCD = 108 degrees (exterior angle of a triangle = sum of interior opposite angles)

I could not proceed any further beyond this. I thought line DB (median) is bisecting angle ADC since line DB is bisecting line AC (AB=BC) but this only happens in case of isosceles and equilateral triangles.

I am very much stuck here and seek your kind suggestions here. I am also suspecting something could be wrong in the diagram of the triangle or may be the unknown angles. Let me know where I am wrong. Thanks in advance.

Answer 1

Hint- Using Law of Sines-

$$\frac{CD}{\sin\angle A}=\frac{AC (=2CD)}{\sin\angle ADC(=72^{\circ})} \implies \sin\angle A=\frac{\sin72^{\circ}}{2}$$

Hope this helps!!

Answer 2

Can you use trigonometry?

By the law of sines, and letting $\angle CAD = \theta$,

$\frac{\sin \theta}{1} = \frac {\sin 72^{\circ}}{2}$

$\sin \theta = \frac 12\sin 72^{\circ}$

I'm pretty sure trigonometry is needed here because the answer is not "nice" (in degrees at least).