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

 A: 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$
$\implies x = \sqrt5$
A: (Lengthy method)$$\angle EDA,  \angle ADB, \angle BDC= 36^{\circ}$$ Since, $ADB$ is isosceles you can find $\angle ABD$ as well and then $\angle TBK$. You may see $\angle TBK=\angle BCD$. Let the side length of pentagon be $a$ and find $BC$ in terms of $a$ and apply cosine rule to $TBK$ and $BCD$ to find $a$. Then you can find $CH$.
A: 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
$$\begin{align*}
CH^2 &= CD ^2 + DH^2 - 2 CD \cdot ND\\
&= s^2 + \left(\frac{\varphi s}{2}\right)^2 - 2s\cdot \frac{s}{4}\\
&= \left[1 + \frac 14\cdot \frac{3 + \sqrt 5}{2} - \frac 12 \right]s^2\\
&= \frac {7 + \sqrt 5}8\cdot \frac{40}{7 + \sqrt 5}\\
&= 5\\
CH &= \sqrt 5
\end{align*}$$
A: I will compute $BH$ (instead, since we can forget about the point $C$ now).
The following solution is purely geometric, and generalizes the given situation.
All we need from the regular pentagon $ABCDE$ is the starting triangle $ABD$, which can be arbitrary.
(No need for the special situation of an isosceles triangle with very particular angles.)

The generalization works in the following picture.


In the given picture show that $BH$ is half $TK$.

Let us state explicitly:

Proposition: The general triangle $\Delta ABD$ is given.
On its sides $AB$ and $BD$ we construct in exterior the squares $ABTU$ and $BDKJ$. Let $L,H,S$ be respectively the mid points of $UJ$, $AD$, $TK$.
Then:
$\bf(1)$ The triangles $\Delta LAD$ and $\Delta LTK$ are isosceles, each with a right angle in $L$.
$\bf(2)$ $BHLS$ is a parallelogram.
$\bf(3)$ (Bonus) $ADKT$ is an
orthodiagonal quadrilateral, i.e. its diagonals $AK$ and $TD$
intersect orthogonally in a point, let it be denoted by $X$.
Then $X$ is also on the lines $UJ$ and $ZY$, and on the circumcircles of the for squares
from the picture.

Note that in the given constellation of points, we have more instances of the "same problem". As stated, $\Delta ABD$ "produces" the dual triangle $\Delta TBK$,
and the points $L,H,S,X,Y,Z$. But we may also start with an other triangle, realizing a constelation with points already knwon.
For instance:
$$
\begin{array}{|l|l||c|c|c|c|c|c|}
\hline
\text{Triangle} & \text{Dual} & L & H & S & Y & Z & X\\\hline
\Delta ABD &\Delta TBK & L & H & S & Y & Z & X\\\hline
\Delta TBK &\Delta ABD & L & S & H & Z & Y & X\\\hline
\Delta TLA &\Delta KLD & B & \text{mid }TA & \text{mid }KD & U & J & X\\\hline
\Delta KLD &\Delta TLA & B & \text{mid }KD & \text{mid }TA & J & U & X\\\hline
\end{array}
$$

Before we show the proposition, let us observe that $(1)$ and $(2)$ easily solve the problem in the OP, because of
$$
BH = SL = ST=SK=\frac 12TK \ .
$$

Proof of the proposition:

$(1)$
Let us start with $A,D$, and build the isosceles triangle $\Delta A\Lambda D$ with a right angle in $\Lambda$. (In the end, we will have $\Lambda =L$.)
We consider the following composition of linear isometries of the plane, applied on a point $W$. Start with $W$,
rotate it (in trigonometric sense) $90^\circ$ around $A$,
take the symmetric point of the result w.r.t. $\Lambda$,
rotate this second result in trigonometric sense $90^\circ$ around $D$.
It is easy to see that the composition is a linear isometry, fixes $A$, $\Lambda$, $D$, so it is the identity.
We apply this transformation on $B$, so $B\to B$. How is this result obtained?
The first step moves $B\to U$,
and the third step moves $J\to B$, so the second step is $U\to J$, a reflection in $\Lambda$.
This shows $L=\Lambda$ is the mid point of $UJ$. So $\Delta ALD=\Delta A\Lambda D$ is isosceles, with right angle in $L=\Lambda$.
The same argument for the "dual" triangle $\Delta TBK$ shows $\Delta TLK$ isosceles, with a right angle in $L$, since the "new point $\Lambda$" (built for $\Delta TBK$) is also characterized by being the mid point of $UJ$.
(In this "duality", $U,J$ are self-dual.) This concludes $(1)$.

A second simpler proof for $(1)$ uses complex numbers.
The lower case letter (variable) associated to a capital one (denoting a point) is the affix of the point (a complex number).
We may and do place $H$ in the origin $0$, so $h=0$. Then $a+d=0$, so $d=-a$. We compute now in terms of $a,b$ all the needed points. We have $u= a+i(b-a)$, $j=d-i(b-d)=-a-i(b+a)$, so $l=\frac 12(u+j)=\frac  12(i(b-a)-i(b+a))=-ia=id$, so $L$ is indeed obtained by a $90^\circ$-rotation of $D$ around $H$, and/or by a $-90^\circ$-rotation of $A$ around $H$.

$(2)$
A proof of $(2)$ using complex numbers is straightforward. For my taste, i am placing the origin of the complex plane in $b=0$.
Then the four points $D$, $S$, $L$, $H$ are easily computed in terms of $a,d$:
$$
\begin{aligned}
h &= \frac 12(a+d)\ ,\\
s &= \frac 12(t+k)=\frac 12(-ia+id)=\frac i2(d-a)\ ,\\
l &= h+ i(d-h) =h+\frac i2(d-a)=h+s\ .
\end{aligned}
$$
So the mid point of $SH$, and the mid point of $BL$ coincide. (The affixes are $\frac 12(s+h)=\frac 12(b+l)$.)
So $BHLS$ is a parallelogram.

The bonus $(3)$: For diversity, let us use vectors and scalar products. All involved factors below, joined by the dot scalar product are vectors. Then:
$$
\begin{aligned}
AK\cdot DT 
&= (BK-BA)\cdot (BT-BD)\\
&= 
\underbrace{(BK\cdot BT + BA\cdot BD)}_{=0}
- \underbrace{BK\cdot BD}_{=0}
- \underbrace{BA\cdot BT}_{=0} \\
&=0\ .
\end{aligned}
$$
We have used the fact that the scalar product of two vectors depends on the norms of the vectors, and on the angle between them, such that when we pass to the supplementary angle preserving the norms, the scalar product changes sign. Here, $\widehat{ABD}+\widehat{TBK}=180^\circ$,
so the two angles involved in the scalar products in the parenthesis are supplementary.
The scalar product of the vectors $AK$ and $DT$ is zero, so they are orthogonal.
Let us show that $X$ is also on $UJ$. Because of $\widehat{AXD}=90^\circ=\widehat{ALD}$,
the point $X$ is on the circle $ALDY$. With the same argument,
it is on the circumcircles of the other three squares $TLKZ$, $ABTU$, $DBKJ$. So we can compute:
$$
\begin{aligned}
\widehat{UXJ}
&=
\widehat{UXA}+
\widehat{AXD}+
\widehat{DXJ}\\
&=
\widehat{UBA}+
\widehat{AXD}+
\widehat{DLJ}\\
&=
45^\circ + 90^\circ + 45^\circ\\
&=
180^\circ\ ,
\end{aligned}
$$
so the points $U,X,J$ are colinear. With the same argument, $X$ is also on $YZ$.
This finishes the proof of the proposition.
$\square$

Here is a final heavy, inflated, painty picture of the story:


Final bonus: In the conditions of the proposition above, assume that $A,B,D$ are vertices of the
regular pentagon $ABCDE$, the one given in the OP. We construct the squares $ABTU$ and $DBKJ$, and the mid points $S,H$ as in the proposition.
Let $\Sigma$, $\Xi$ be the mid points of $AT$ and $DK$. Let $A'BCD'E'$ and $H'$ be the reflection in $BC$ of the pentagon $ABCDE$, and respectively
of the point $H$. Then we have:
$\bf(1)$ $C$ is also on the line $UXLJ$.
$\bf(2)$ The following lines are related by parallelism:
$$
UXLCJ\ \| \
\Sigma \Xi\ \| \
TD'\ \| \
SH'\ \| \
A'K\ .
$$

This bonus is the subject of a follow up question.
A: See 2nd drawing of the initial post for notations.
CH=BH ( 1 ) by symmetry; a 90 degs clockwise rotation about B sends A to T, T to T', K to D, hence T'D=TK ( 2 ). BH is midline of triangle ADT', so CH=BH=T'D/2=TK/2, done.
