<s>[INST] What is the value of the $CH$ segment in the figure below? For reference: In the figure, $ABCDE$ is a regular pentagon with $BD = BK, AB = BT ~and ~TK = 2\sqrt5$. Calculate $CH$ (If possible by geometry instead of trigonometry)

My progress:

$Draw KD \rightarrow \triangle DBK(isosceles)\\
Draw TAB \rightarrow \triangle BTA(isosceles)\\
a_i = \frac{180(5-2)}{5} = 108^\circ\\
\angle A EH= 360 -2(108)-2(90)=54^\circ$
but I can't finish...
i I made the figure of peterwhy

 [/INST] Given $CH$ is half of $TK$, there should be some clever construction. I am not getting it right now. But here is a solution using trigonometry.
Say $CH = x, DE = a$. If $EH$ meets $BC$ at $M$ then $\angle MEC = 18^0$
$CE = 2 a \cos 36^\circ$
$EM = CE \cos 18^\circ = 2 a \cos 36^\circ \cos18^\circ = a (\cos 18^\circ + \cos 54^\circ)$
$EH = a \cos 54^\circ$
So, $HM = a \cos 18^\circ$. If $\angle CHM = \theta$,
$\tan\theta = \cfrac{CM}{HM} = \cfrac{1}{2 \cos 18^\circ}$
$\sin^2\theta = \cfrac{\tan^2\theta}{1+\tan^2\theta} = \cfrac{1}{3 + 2 \cos 36^\circ} \tag1$
Now $CM = CH \sin\theta \implies a = 2 \ x \sin \theta \tag2$
$BT = a, BK = CE = 2 a \cos 36^\circ$
Note $\angle KBT = 108^\circ$. Applying law of cosine in $\triangle BKT$,
$(2 \sqrt5)^2 = a^2 + 4 a^2 \cos^2 36^\circ - 4 a^2 \cos 36^\circ \cos 108^\circ$
$ = a^2 + 2 a^2 (2 \cos^2 36^\circ + 2 \cos 36^\circ \cos 72^\circ)$
$ = a^2 + 2a^2 (1 + \cos 72^\circ + \cos 36^\circ - \cos 72^\circ)$
$ = a^2 (3 + 2 \cos 36^\circ) = \cfrac{a^2}{\sin^2\theta} = 4 x^2$
$\implies x = \sqrt5$
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