Is it possible to find such an angle using only angle chasing? I've been trying to solve some problem and I came down to the following seemingly easy question:

given two triangles ABC and ABD, and their corresponding angles, how do we find the angle $\angle ACD$ using only angle chasing, I know that it's possible to do that by using coordinate bashing or law of cosine for example, but I was wondering if it's actually possible to do so using only angle chasing, no trigonometry involved.
 A: The answer is: It is impossible with only angle chasing. (It may be possible for some limited special cases.)
In general, the angle $∠ACD$ is related to known angles via the sine equation below,
$$\sin (∠ACD )\sin (∠ACD-∠C) = \frac{\sin ( ∠CAD) \sin (∠CBD) \sin (∠DAB)}{\sin (∠B) }$$
For arbitrary triangles ACB and ADB, the above relationship  can not be reduced to just angle additions and subtractions.
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A well-known geometry problem pertinent to the topic here is the so-called the "hardest easy angle question" below:

A: Here a certain case is considered.
 Draw a circle which passes points A, B, and D. Draw a chord from D parallel with AB , it crosses the circle at E.Connect E to D and B. We have:
$\angle BED=\angle DAB$
$\angle AEB=\angle ADB$
Summing these we get:
$\angle AED=\angle BED+\angle AEB=\angle DAB+\angle ADB$
If the position of C is such that ED  and EC are  the bisector of $\angle CDA$ and $\angle CAD$ respectively then in triangle ACD we have:
$\angle AED=\frac{\angle ACD}{2}+\frac{\pi}{2}$
Therefore we have:
$\angle DAB+\angle ADB=\frac{\angle ACD}{2}+\frac{\pi}{2}$
Or:
$\angle ACD = 2(\angle DAB+\angle ADB)-\pi$
