Thales Theorem : Any triangle inscribed in a circle with one side a diameter is a right angle. I wanted to show Thales Theorem using algebra i.e.,
if $\overrightarrow{\mathbb p} =\overrightarrow{AB},\overrightarrow{\mathbb q} =\overrightarrow{AC},\overrightarrow{\mathbb d} =\overrightarrow{CB} $ I want to show $\overrightarrow{\mathbb q} \cdot \overrightarrow{\mathbb p}=0.$
This is what I did :
It is seen from the figure that $\|\overrightarrow{\mathbb d}\|= \|\text{proj }_{\overrightarrow{\mathbb d}}\overrightarrow{\mathbb p}\|+\|\text{proj }_{\overrightarrow{\mathbb {(-1)d}}}\overrightarrow{\mathbb q}\|$ and $\overrightarrow{\mathbb d} = \overrightarrow{\mathbb p} - \overrightarrow{\mathbb q}$
where $\text{proj }_{\overrightarrow{\mathbb a}}\overrightarrow{\mathbb b}$ is the projection vector of $\overrightarrow{\mathbb b}$ on $\overrightarrow{\mathbb a}$
Thus \begin{align} \|\overrightarrow{\mathbb d}\| &= \frac{\overrightarrow{\mathbb p} \cdot \overrightarrow{\mathbb d}}{\|\overrightarrow{\mathbb d}\|}+\frac{\overrightarrow{\mathbb q} \cdot \overrightarrow{\mathbb {(-1)d}}}{\|\overrightarrow{\mathbb {(-1)d}}\|}\\ \|\overrightarrow{\mathbb d}\| &= \frac{\overrightarrow{\mathbb p} \cdot \overrightarrow{\mathbb {p-q}}}{\|\overrightarrow{\mathbb d}\|}+\frac{\overrightarrow{\mathbb q} \cdot \overrightarrow{\mathbb {(q-p)}}}{\|\overrightarrow{\mathbb {d}}\|}\\ \|\overrightarrow{\mathbb {d}}\|^2 &=\|\overrightarrow{\mathbb p}\|^2 + \|\overrightarrow{\mathbb q}\|^2 - 2 \overrightarrow{\mathbb p} \cdot \overrightarrow{\mathbb q} \end{align}
Now consider the vector $\overrightarrow{\mathbb d'}= \overrightarrow{AD}$ which passes through the center so that $\overrightarrow{\mathbb d'}$ is also a diameter.
Since $\overrightarrow{\mathbb d'}$ and $\overrightarrow{\mathbb d}$ bisect each other $ABDC$ form a parallelogram and hence $\overrightarrow{\mathbb d'}= \overrightarrow{\mathbb p}+ \overrightarrow{\mathbb q}$ which implies $$\|\overrightarrow{\mathbb d'}\|^2 =\|\overrightarrow{\mathbb p}\|^2 + \|\overrightarrow{\mathbb q}\|^2 + 2 \overrightarrow{\mathbb p} \cdot \overrightarrow{\mathbb q} $$
Since $\|\overrightarrow{\mathbb d'}\|^2 = \|\overrightarrow{\mathbb d}\|^2,$ we have $\overrightarrow{\mathbb p} \cdot \overrightarrow{\mathbb q} = - \overrightarrow{\mathbb p} \cdot \overrightarrow{\mathbb q} $ and thus $$\overrightarrow{\mathbb p} \cdot \overrightarrow{\mathbb q} =0$$
Question How can I justify that I should take the projection of $\overrightarrow{\mathbb q}$ along $\overrightarrow{\mathbb {(-1)d}}$ and not $\overrightarrow{\mathbb d}?$
Also my proof looks tedious, any simpler way to do it?
You can shorten your proof, and bypass your concerns about projection, by arguing directly: $$ \|d\|^2=d\cdot d = (p-q)\cdot(p-q)=\|p\|^2 +\|q\|^2 -2p\cdot q $$ using properties of the inner product.