<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] The following is based on the property of regular pentagons, that the ratio of their diagonals to their sides are in golden ratio $\varphi = \frac{1 + \sqrt 5}{2}$. For example,
$$\frac{BK}{BT} = \frac{BD}{BA} = \varphi$$
Rotate $\triangle KBT$ about $B$ clockwise by $90^\circ$ to $\triangle DBT'$. Then $ABT'$ is a straight line. Because $\triangle ABD$ is isosceles, drop altitude $DM$ onto $AB$, and $M$ would be the midpoint of $AB$. (Diagram by OP peta arantes)

Let $s$ be the side length of the regular pentagon. Then consider $\triangle DBT'$ and the foot of altitude $M$,
$$\begin{align*}
DT'^2 &= BT'^2 + BD^2 + 2 BT'\cdot BM\\
(2\sqrt 5)^2 &= s^2 + (\varphi s)^2 + 2s \cdot \frac s2\\
20 &= s^2 + \frac{3 + \sqrt 5}{2} s^2 + s^2\\
&= \frac{7 + \sqrt 5}{2} s^2\\
s^2 &= \frac{40}{7 + \sqrt 5}
\end{align*}$$
Next, consider $\triangle CDH$. Drop altitude $HN$ onto side $CD$, where $N$ is the foot of the altitude. Since $\triangle CDA$ is isosceles and $DH = \frac 12 DA$, so $ND = \frac 14 CD = \frac 14 s$.
The required length $CH$ can be found by
$$\begin{align*}
CH^2 &= CD ^2 + DH^2 - 2 CD \cdot ND\\
&= s^2 + \left(\frac{\varphi s}{2}\right)^2 - 2s\cdot \frac{s}{4}\\
&= \left[1 + \frac 14\cdot \frac{3 + \sqrt 5}{2} - \frac 12 \right]s^2\\
&= \frac {7 + \sqrt 5}8\cdot \frac{40}{7 + \sqrt 5}\\
&= 5\\
CH &= \sqrt 5
\end{align*}$$
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