Proof of $a = v^2/r$ using similar triangles 
The book states that the ‘change in velocity’ triangle and the displacement triangle has the same angle theta.
But I don’t get it? How can we prove that the two triangles will have the same angle?
 A: The displacement and velocity vectors in circular motion are always perpendicular:
$$\angle \vec{r}_1 - \angle \vec{v}_1 = 90^\circ \quad \text{and} \quad \angle \vec{r}_2 - \angle \vec{v}_2 = 90^\circ$$
Subtract the above two equations:
$$\angle \vec{r}_1 - \angle \vec{r}_2 - \angle \vec{v}_1 + \angle \vec{v}_2 = 90^\circ - 90^\circ$$
With $\angle \vec{r}_1 - \angle \vec{r}_2 = \theta$ the above equation becomes:
$$\boxed{\angle \vec{v}_1 - \angle \vec{v}_2 = \theta}$$
Assuming that the circular motion is uniform (constant speed)
$$|\vec{r}_1| = |\vec{r}_2| = R, \qquad |\vec{v}_1| = |\vec{v}_2| = v_0$$
the two triangles are similar:
$$\frac{|\Delta\vec{v}|}{|\vec{v}_1|} = \frac{|\Delta\vec{r}|}{|\vec{r}_1|} \qquad \rightarrow \qquad |\Delta\vec{v}| = \frac{v_0}{R} |\Delta \vec{r}|$$
where $\Delta \vec{r} = \vec{r}_2 - \vec{r}_1$ and $\Delta \vec{v} = \vec{v}_2 - \vec{v}_1$. The average acceleration is defined as
$$a_{av} = \frac{|\Delta \vec{v}|}{\Delta t} = \frac{v_0}{R} \frac{|\Delta \vec{r}|}{\Delta t}$$
If we let the time difference go to zero $\Delta t \rightarrow 0$, the average acceleration becomes instantaneous acceleration:
$$a = \lim_{\Delta t \rightarrow 0} a_{av} = \frac{v_0}{R} \lim_{\Delta t \rightarrow 0} \frac{|\Delta \vec{r}|}{\Delta t} = \frac{v_0^2}{R}$$
The acceleration vector is perpendicular to the velocity vector and is also known as radial acceleration or centripetal acceleration:
$$\boxed{a_{\perp} = \frac{v_0^2}{R}}$$
When circular motion is not uniform, the expression for radial acceleration is the same, but there is also tangential acceleration component which is parallel to the velocity vector:
$$\boxed{a_{||} = \frac{d}{dt}v_0}$$
Here you can find detailed derivation of the expressions for radial and tangential acceleration components: https://physics.stackexchange.com/a/685423/149541

Here I give easy-to-understand proof that the displacement and velocity vectors are always perpendicular.
The position in circular motion is defined as
$$x^2 + y^2 = R^2$$
where the radius $R$ is assumed to be constant. Take the time derivative of the above equation:
$$2 x \dot{x} + 2 y \dot{y} = 0$$
With $v_x = \dot{x}$ and $v_y = \dot{y}$ the above equation becomes:
$$x v_x + y v_y = 0$$
Now recognize that the above equation is scalar product between the displacement $\vec{r}$ and velocity $\vec{v}$ vectors:
$$\vec{r} \cdot \vec{v} = |\vec{r}||\vec{v}| \cos \phi = 0$$
The solution to the above equation is $\phi = 90^\circ$ which concludes the proof that displacement and velocity vectors are always perpendicular in circular motion.
A: The direction of vector $v_1$ is the same as vector $r_1$ but rotated 90$^\circ$ degrees clockwise and similarly vector $v_2$ is rotated $90^\circ$ with respect to vector $r_2$. So the angle between vectors $v_1$ and $v_2$ is the same as the angle between $r_1$ and $r_2$. More precisely, if you define $\text{angle}(v_1)$ as the angle that $v_1$ makes with the x-axis (or some other arbitrary axis) and $\text{angle}(v_1,v_2) $ as the angle between $v_1$ and $v_2$ then we get
\begin{align}
\text{angle}(v_1,v_2)&=\text{angle}(v_1)-\text{angle}(v_2)\\
&=\text{angle}(r_1)-90^\circ-\big(\text{angle}(r_2) - 90^\circ\big)\\
&=\text{angle}(r_1)-\text{angle}(r_2)\\
&=\text{angle}(r_1, r_2)
\end{align}
Note that we know that the angle between $r$ and $v$ is always $90^\circ$, otherwise the length of $r$ would be increasing and we wouldn't have circular motion.
A: Okay, so I was working on this question and I think I solved it. Please tell if I made any inaccurate assumptions.

Firstly, we extend the -VA (which originates at the tip of VB, or in this case, E) till it intersects the line OB at D.
Then, we extend OB till it intersects VA at C.
Since, ED (or -VA) is parallel to AC (or VA), angle(ACO) = angle(BDE).
Also, angle(CAO) = angle(DBE) = 90.
Therefore, angle(BED) = angle(AOC) = .
