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I've been trying to solve some problem and I came down to the following seemingly easy question:

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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.

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  • $\begingroup$ No I don't, that only allows you to know the sum $\angle BCD+\angle ADC$ $\endgroup$ – BashForLife Aug 23 at 17:57
  • $\begingroup$ Sorry but I don't see how that can be done, could you please elaborate more $\endgroup$ – BashForLife Aug 27 at 12:16
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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:

enter image description here

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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:

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

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  • $\begingroup$ From your solution it turns out that $\angle ACD$ does not depend on the position of $C$... I suspect something went wrong somewhere... $\endgroup$ – Aretino Sep 1 at 12:58
  • $\begingroup$ @Aretino I suspect this step, which seems to come from nowhere: $\angle AED=\frac{\angle ACD}{2}+\frac{\pi}{2}$. $\endgroup$ – David K Sep 1 at 13:26

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