How to determine arc measures from angles between secant and tangents (without trigonometry) 
Given a circle, a point $H$ outside the circle, segments $\overline{HE}$ and $\overline{HT}$ tangent to the circle at $E$ and $T$, respectively, and points $I$ and $G$ on the circle such that $I$, $G$, and $H$ are collinear (all as shown above), knowing the measures of $\angle EHG$ and $\angle GHT$ (call them $\alpha$ and $\beta$, respectively, for convenience) determines the measures of each of the four arcs on the circle.
Is it possible to compute the measures of the minor arcs $EI$, $IT$, $TG$, and $GE$ in terms of $\alpha$ and $\beta$ without using trigonometry?
 A: I'm within one, so I hope it prompts your thinking.  Basic geometry was a lot of years ago.  The theorems I remember are $\stackrel{\frown}{EGT} = 2\angle EIT$, $\beta=(\stackrel{\frown}{IT} - \stackrel{\frown}{TG})/2$,  and $ET=\pi-\alpha-\beta$, which you can see by drawing radii from E and T and the line from H to the center of the circle.  Then if $\angle EIG=\gamma$ we have $EG=2\gamma, GT=\pi-\alpha-\beta-2\gamma, TI=\pi-\alpha+\beta-2\gamma, IE=2\gamma+2\alpha$
A: With trig, I get that the arcs have measure $\theta \pm \alpha$ and $\pi - \theta \pm \beta$, where 
$$\cos\theta = \frac{\sin\frac{\beta-\alpha}{2}}{\sin\frac{\alpha+\beta}{2}} = \frac{\cos\beta-\cos\alpha}{1-\cos\left(\alpha+\beta\right)}$$
By the "far arc / near arc" principle, $\stackrel{\frown}{EI}-\stackrel{\frown}{EG} = 2\alpha$, so we can write $\stackrel{\frown}{EI} = \theta+\alpha$ and $\stackrel{\frown}{EG}=\theta-\alpha$.
Let's go analytical ... 
Let the circle to have radius $1$, with its center, $O$, at the origin; and let $H$ lie on the positive $x$-axis. In quadrilateral $OEHT$, angles at $E$ and $T$ are right angles, and the angle at $H$ has measure $\alpha+\beta$, so that the angle at $O$ has measure $2 \gamma := \pi-(\alpha+\beta)$. Thus, $\angle HOE = \gamma$ and $|OH| = \sec\gamma$. Moving around the circle by arcs of $\theta \pm \alpha$ from $E$ gets us to $I$ and $G$, and we determine coordinates for these points, writing "$\rm{cis}\cdot$" for "$(\cos\cdot,\sin\cdot)$":
$$I = \rm{cis}\left(\gamma + ( \theta + \alpha )\right) = \rm{cis}\left(\gamma+\alpha+\theta\right)$$ 
$$G = \rm{cis}\left(\gamma-(\theta-\alpha)\right) = \rm{cis}\left(\gamma+\alpha-\theta\right)$$
Since $H$, $G$, and $I$ are collinear, we must have equal slopes for segments $HI$ and $HG$:
$$\frac{ \sin(\gamma+\alpha+\theta) }{ \cos(\gamma+\alpha+\theta)-\sec\gamma } = \frac{ \sin(\gamma+\alpha-\theta) }{ \cos(\gamma+\alpha-\theta)-\sec\gamma }$$
Therefore,
$$\sin(\gamma+\alpha+\theta)\left(\cos(\gamma+\alpha-\theta)-\sec\gamma\right) = \sin(\gamma+\alpha-\theta) \left( \cos(\gamma+\alpha+\theta)-\sec\gamma\right) $$
$$\begin{eqnarray*}
&&\sin(\gamma+\alpha+\theta)\cos(\gamma+\alpha-\theta)- \sin(\gamma+\alpha-\theta) \cos(\gamma+\alpha+\theta)  \\
&=& \sec{\gamma} \; \left( \sin(\gamma+\alpha+\theta) -\sin(\gamma+\alpha-\theta) \right) \\
\sin 2\theta &=& 2 \sec{\gamma} \; \sin \theta \cos\left(\gamma+\alpha\right) \\
2 \sin\theta \cos\theta &=& 2 \sec{\gamma} \; \sin \theta \cos\left(\gamma+\alpha\right)
\end{eqnarray*}$$
If $\theta\ne 0$ and $\theta \ne \pi$ [*], then we can cancel $\sin\theta$ and finish-up:
$$\begin{eqnarray*}
\cos\theta &=& \frac{\cos\left(\gamma+\alpha\right)}{\cos\gamma} = \frac{\cos\frac{\pi+\alpha-\beta}{2}}{\cos\frac{\pi-(\alpha+\beta)}{2}}= \frac{\sin\frac{\beta-\alpha}{2}}{\sin\frac{\alpha+\beta}{2}} \\
\end{eqnarray*}$$
There's certainly a way to arrive at this result without using coordinates and slopes (I got lazy), but if there's a way to get there "without using trigonometry", I'm not seeing it.
Edit.
I'll just point out that, if $F$ is the foot of the perpendicular from the circle's center to segment $HI$, then
$$\frac{|OF|}{|OE|}=\frac{|OH| \sin\angle FHO}{|OH|\sin\angle EHO}=\frac{ \sin\frac{\alpha-\beta}{2}}{\sin\frac{\alpha+\beta}{2}}=\cos\left(\pi-\theta\right)$$
Also, there are points $P$ and $Q$ on the circle, with $P$ on $\stackrel{\frown}{IE}$ and $G$ on $\stackrel{\frown}{QE}$, such that $\stackrel{\frown}{IP} = \stackrel{\frown}{GQ} = \alpha$, whence $\stackrel{\frown}{PE}=\stackrel{\frown}{QE}=\theta$. So, all the components of the final relation appear in the figure somewhere. I haven't looked hard enough to determine if it's possible to "see" the relation geometrically.
End edit.
[*] If $\theta = 0$, then $\alpha$ itself must be $0$ (so that $\theta-\alpha$ is a non-negative arc measure), with $I$ and $G$ coinciding with $E$. Likewise, $\theta = \pi$ implies $\beta = 0$, with $I$ and $G$ coinciding with $T$. Analysis of these cases is straightforward, with the same formula resulting as in the general case.
A: There is no solution without inverse trigonometric functions, because any elementary geometric argument would, for a given value of the angle between the tangent lines, calculate all 4 arc lengths as linear functions of $\alpha$.  This would imply that, as the (secant) line through $H$ is rotated at uniform angular speed, it sweeps out arclength on the circle at a uniform rate.  But this is impossible because the distance from $H$ to the intersection point, $|HG|$, is not constant.  It is not hard to make this argument rigorous using calculus.
If the arcs, in the order listed in the question, are denoted $A, B, b, a$, then  elementary geometry does express $(A-a)$, $(B-b)$, and $(A+B+b+a)$ as linear functions of $\alpha, \beta$ and $1$, as in the other solutions.   I think this set of 3 equations in 4 unknowns is what the OP meant by "not enough equations" to solve everything purely geometrically.
