I will try to explain the solution:
QUESTIONS: 1. Why is the elongation distribution in the length of the beam a triangle?
The "real-life" visualization of the displacement of the system is supposed to be circular. In the picture, the red lines indicates the exaggerated actual movement of points in the system. As depicted, the end of beam, C', displaces by $x_2$ and $y_2$.
To answer your question, the relationship between the original position and the deflected position of the rigid beam lies in the concept of trigonometry. In civil engineering, we design structures to avoid excessive deflections and angular rotations, such that the angle of depression on the beam, denoted by B in the figure, is very small, e.g. B < 5 degrees. Therefore, the expected _vertical displacement, $y_2$ of the system is relatively small compared to the length of the beam itself. Also, it is important to note that since $y_2$ is small, $x_2$ will also be extremely small, which is close to a negligible value.
To provide an easier analysis, we therefore then assume that the deflected beam projects a triangular profile because C' is almost vertically collinear with C.
Note: $X_2 = Y_2\tan(B)$
Question 2. Why is the elongation $𝛿_{st}$ computed as the elongation in the cable
The explanation goes similar with question 1, but this does not include arcs. The simplest I can get is that: For small angles, say angle A, the hypotenuse of a triangle is approximately equal to one of its leg.
Thus, the position of the end of the cable after displacement does not matter anymore as it is very close to zero.
The above explanations makes engineering simpler. Take note that using that concept, you can easily get the displacement of point C just by ratio and proportion.