geometryphysics
This is a common setup for kinematics problems in physics. My geometry is rusty and I want to understand this very simple idea.
I am having trouble understanding why the angle $\theta$ formed by $\overrightarrow{w}$ is equal to $\theta$ = $\angle$ BOA.
My initial ideas:
According to the Wikipedia article absolute geometry the first 28 propositions of Euclid's Elements do not require the parallel postulate and hence may be proven for neutral or "absolute" geometry.
Prop. 11 is the construction of a perpendicular to a line from a point on the line. Prop. 12 is the construction of a line perpendicular to a given line through a point not on the line. So, combined with OP's proof that all lines perpendicular to a given line are parallel, the result is established: The family of parallel lines perpendicular to a given line covers all points.
in $ \triangle ABC$, $\angle B=90^\circ$, $AD=DC$
Best Answer
My answer is essentially the same as the one given by half-integer fan, but I'll add a picture in case it will aid your understanding. I have labeled new points $C$ and $D$ so that I can refer to them.
$\triangle OCD$ is a right triangle, as is $\triangle OCE$. Because $$\angle OCD=90^\circ=\angle OCE+\angle DCE=\angle OCE+\alpha$$ and $$\angle OCE+\angle COE+90^\circ=\angle OCE+\theta+90^\circ=180^\circ$$ we have that $$\angle OCE +\theta=90^\circ\quad \text{ and }\quad \angle OCE+\alpha=90^\circ,$$ so we must have $\theta=\alpha$.