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I have a quadrilateral ABCD. If I know the angles $DAB, DCB, ADB, CDB, ABD, CBD$, can I obtain the angle $DAC$? I know that I can do it by some coordinate geometry or the sine formula, but is there a more elegant way?

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  • $\begingroup$ Can you explicitly show what you already know about how to solve this problem ? $\endgroup$
    – disgraced
    Commented May 16, 2022 at 9:17

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If you need general answer for any values of angles there is no way to avoid trigonometric functions.

Let $\angle ADB=\alpha$, $\angle ABD=\beta$, $\angle CDB=\gamma$, $\angle CBD=\delta$, $\angle DAC=x$.

Then $\angle DAB=\pi-(\alpha+\beta)$, $\angle DCA=\pi-(\alpha+\gamma+x)$, $\angle DCB=\pi-(\gamma+\delta)$.

Let mark for convenience $AD=a\sin\beta \sin(\gamma+\delta)$. Then using sine law:

$$BD=a\sin(\alpha+\beta)\sin(\gamma+\delta), CD=a\sin\delta \sin(\alpha+\beta)$$ $$AD\sin DAC=CD\sin DCA\Rightarrow \sin\beta \sin(\gamma+\delta)\sin x=\sin\delta \sin(\alpha+\beta)\sin(\alpha+\gamma+x)$$ $$\sin\beta\sin(\gamma+\delta)\sin x=\sin\delta \sin(\alpha+\beta)(\sin(\alpha+\gamma)\cos x+\cos(\alpha+\gamma)\sin x)$$ $$\tan x=\frac{\sin\delta \sin(\alpha+\beta)\sin(\alpha+\gamma)}{\sin\beta\sin(\gamma+\delta)-\sin\delta \sin(\alpha+\beta)\cos(\alpha+\gamma)}$$

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