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.