Physics equations/Geometrical discussion of uniform circular motion

Using the figure we define the distance traveled by a particle during a brief time interval, $\Delta t$, and the (vector) change in velocity:

uniform circular motion (here the Latin d was used instead of the Greek Δ

1     $\Delta \ell =|{\vec {r}}_{2}-{\vec {r}}_{1}|$, and $\Delta v=|{\vec {v}}_{2}-{\vec {v}}_{1}|$

2     $\Delta \ell =v\Delta t$ (rate times time equals distance).

3     $\Delta {\vec {v}}={\vec {a}}\Delta t$ (definition of acceleration).

4     $\Delta v=a\Delta t$ (taking the absolute value of both sides).

5     ${\frac {\Delta v}{v}}={\frac {\Delta \ell }{r}}$ (by similar triangles). Substituting (2) and (4) yields:

6     ${\frac {a\Delta t}{v}}={\frac {v\Delta t}{r}}$, which leads to ${\frac {a}{v}}={\frac {v}{r}}$, and therefore:

7     $a={\frac {v^{2}}{r}}$