Why are these triangles similar ? (Derivation of uniform circular motion equation ) I am studying the equations of uniform circular motion and I am  having difficulties  with the derivation of them.

I quote my book :


  We can find a simple expressio for the magnitude of the acceleration
  in uniform circular motion.We begin with the below diagram,which shows
  a particle moving with constant speed in a circular path of radius $R$
  with center at $O$ . The particle moves a distance $ \Delta s$ from
  $P_1$ to $P_2$ in a time interval $\Delta t$. The figure below also
  shows the vector change in velocity $\Delta \vec{v}$ during this
  interval.  The angles labeled $\Delta \phi $ in the figure are the
  same because $\vec {v_1}$ is perpendicular to the line $OP_1$ and
  $\vec{v_2}$ is perpendicular to the line $OP_2$. Hence the triangle in
  the figure are similar.
  


I sadly don't see how the angles are equal and how the triangles are similar as a consequence of the fact that $\vec {v_1}$ and $\vec {v_2}$ are perpendicular to $R$.

Can you guys help with a math proof ?
Note that $v_1$ and $v_2$ are equal in magnitude and perpendicular to the respective radii. So they form an isosceles triangle and the angle between them is the same as the angle between the radii. The base angles of the two isosceles triangles are also equal, hence they are similar.



To show the angles are equal translate the configuration at $P_2$ to $P_1$ and work round the angles in that figure. The angle between $v_2$ and $OP_1$ adds to the angle between $OP_2$ and $OP_1$ to give a right angle and similarly with the angle between $v_2$ and $v_1$ so the angles you want are equal.
What helped me is this: If the line $O-P_1$ was rotated by 90 degrees, you would be able to see that the difference between $v_1$ and $v_2$ would also be 90 degrees.

This holds true when the vectors flip 180 degrees as well, so logically (to me at least) it should hold true for any angle between $O-P_1$ and $O-P_2$.