Why are angles $\alpha_0$ and $\alpha_2$ equal? Can someone please tell me what rule in geometry would show why angles  $\alpha_0$ and  $\alpha_2$ are equal?  I can see that angles a0 and a1 are equal because the red line is acting as a transversal line.  But I'm stumped as to what rule proves  $\alpha_0$0 is equal to  $\alpha_2$.

 A: Perhaps a longer walk through would be helpful, though the comments contain a good answer already.
You already know that $\alpha_0 = \alpha_1.$ So let's show that $\alpha_1 = \alpha_2$ and call it a day. If we think about the half circle of which $\alpha_1$ is a part, we know the angles here must add up to $180^\circ$ (or $\pi$ radians). We also know that the rightmost angle here must have size $90^\circ- \alpha_2,$ because it's part of a right triangle with $\alpha_2.$
Symbolically we have:
$$\alpha_1 + 90^\circ + (90^\circ - \alpha_2) = 180^\circ$$
or
$$\alpha_1 - \alpha_2 = 0$$
Which is exactly what we wanted. Hence $\alpha_0 = \alpha_1 = \alpha_2$
A: 
Rotate the hinge through $90^\circ$ and you get angles on parallel lines.
A: In the picture we see that the area with two pink and two red sides is a quadrilateral which means that the sum of all its angles is 360° .
Also two angles of the quadrilateral are 90° each i.e. their sum is 180° .
So the sum of rest of the two angles is also 180° .
Again the adjacent angle of alpha 2 is : 180° - alpha 2 (by linear pair of angles).
Now a1 + 180°-a2=180°
Hence a1 = a2.
(a here refers to alpha)
