Is it possible to find such an angle using only angle chasing?

Question

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.

Answer 1

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:

Answer 2

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:

$\angle ACD = 2(\angle DAB+\angle ADB)-\pi$