what is the value of angle $a$

Question

what is the value of angle $a$

Answer 1

(We are grateful to Suleyman Soyler for this solution, he is a math teacher.)

By power of $D$ w.r.t given circle: $|DC|^2=|DE|\cdot |DB|$ and $|AD|=|DC|$. Therefore

$$|AD|^2=|DE|\cdot |DB|$$

implies $\triangle ADB \sim \triangle EDA$ (by side-angle-side). Hence we find that

$$ \angle ABD = \angle EAD \tag{1}$$

$$ \angle DEA = \angle DAB = 36^\circ \tag{2}$$

Now, let's draw circumcircle of $\triangle AED$ and ray $[DF$. Let's say intersection of them $K$. By inscribed angles

$$ \angle EAD = \angle EKD \tag{3}$$

and $$ \angle DKA = \angle DEA = 36^\circ \tag{4}$$

By sum of interior angles of $\triangle ADK$, we find $\angle KAF =36^\circ$ and therefore $|FB|=|FA|=|FK|$. Then

$$\angle FBK = \angle FKB = 54^\circ \tag{5}$$

On the other hand, $(1)$ and $(3)$ implies that $KFEB$ a cyclic quadrilateral. Thus $$ \angle DEF = \angle FKB = 54^\circ $$

Answer 2

Answer: $\angle FED = \alpha = 54^\circ .$ In the solution, we will use $\cos 54^\circ = \sqrt{\dfrac{5-\sqrt 5}{8}} $ and $\cos 36^\circ = \dfrac{\sqrt 5 -1}{4}$.

Without loss of generality we assume that $|AD|=|AF|=2.$ By $\cos 36^\circ = \dfrac{\sqrt 5 -1}{4}$ and cosine theorem in $\triangle ADF$, we find $|DF|=\sqrt 5 -1.$ Let's say that $|DE|=x, |EB|=y, |FE|=z.$ By power of point $D$ with respect to the circle:

$$x(x+y)=2^2 \tag{1}$$

On the other hand, by cosine theorem in $\triangle ADB$: $(x+y)^2=2^2+4^2-2\cdot 2\cdot 4 \cdot \cos 36^\circ$ and

$$ (x+y)^2=16-4\sqrt 5 \tag{2} $$

If we solve $(1)$ and $(2)$ together we can find

$$x=\dfrac 2{\sqrt{4 - \sqrt5}}, \ \ \ y=\dfrac {2(3-\sqrt 5)}{\sqrt{4 - \sqrt5}} \tag{3} $$

Now, let's apply Stewart's theorem in $\triangle BDF$: $z^2 =\dfrac{x\cdot |BF|^2 + y\cdot |DF|^2}{x+y}-xy$ and we can calculate $z$, by exhausting operations.

In last step, $\angle DEF = \alpha$ and $\cos\alpha = \dfrac{x^2+z^2-|DF|^2}{2 x z}$ and therefore $\cos \alpha = \sqrt{\dfrac{5-\sqrt 5}{8}} \implies \alpha = 54^\circ $.