Geometrically deriving Lorentz transformation from Minkowski diagram How can we derive Lorentz transformation from a Minkowski diagram (like below image) by using only geometry theorems such as sines theorem and Pythagoras theorem?

 A: It can be done. But if the Lorentz diagram is taken in the literal Euclidian sense the transformations obtained read
$$
x = \frac{1}{\sqrt{1+\beta^2}}\left(x' + \beta ct'\right)\\
ct = \frac{1}{\sqrt{1+\beta^2}}\left(ct' + \beta x'\right)
$$ 
with $\beta = v/c$. The reason is not hard to see: the Euclidian distance on the diagram is $\sqrt{(ct)^2 + x^2}$, whereas the invariant space-time interval is $\sqrt{(ct)^2 - x^2}$. 
If you insist on a purely geometrical derivation, one remedy would be to observe that the $x'$ and $ct'$ axes must be assigned different scaling than the $x$ and $ct$ axes. That is, things are kept consistent if the primed units are required to be $\sqrt{\frac{1+\beta^2}{1-\beta^2}}$ of the unprimed ones. As long as this scaling is applied, geometry and trigonometry may be carried on as usual. But then the scaling must be justified beforehand, and this would mean referring to either time dilation or length contraction. 
The better alternative is to account for the hyperbolic nature of the space-time distance from the beginning. This amounts to replacing trigonometric functions by their hyperbolic counterparts. The result is that the factor of $\frac{1}{\sqrt{1+\beta^2}}$ in the transformations above is replaced by the usual $\frac{1}{\sqrt{1-\beta^2}}$.
This being said, the idea is to write $x$ and $ct$ in terms of segments that can be easily related to $x'$ and $ct'$. To this end, denote $\alpha$ the angle between the $ct$ and $ct'$ axes. Project the coordinate point on the $ct'$ axis onto the $ct$ axis, then extend the projection onto the normal from event E to the $x$ axis. 

Now use the two right triangles so formed to notice that, geometrically, we can write
$$
x = x' \cos\alpha + ct' \sin\alpha \\
ct = x' \sin\alpha + ct' \cos\alpha
$$
If we go the Euclidian way, then $\tan\alpha = \beta$, $\cos\alpha = \frac{1}{\sqrt{1+\beta^2}}$, $\sin\alpha = \frac{\beta}{\sqrt{1+\beta^2}}$ and the scaling of the $x'$ and $ct'$ axes requires $x' \rightarrow x'\sqrt{\frac{1+\beta^2}{1-\beta^2}}$, $ct' \rightarrow ct' \sqrt{\frac{1+\beta^2}{1-\beta^2}}$. 
If we account for the space-time interval as invariant "distance", then $\tan\alpha \rightarrow \tanh\alpha = \beta$, $\cos\alpha \rightarrow \cosh \alpha = \frac{1}{\sqrt{1-\beta^2}}$, $\sin\alpha \rightarrow \sinh\alpha = \frac{\beta}{\sqrt{1-\beta^2}}$, and no scaling need be applied.
