<s>[INST] Showing point is the orthocenter Given a rectangle WXYZ, let R be a point on its circumscribed circle. Show that, out of the orthogonal projections of R onto WX, XY, YZ, and ZW; one out of these 4 points is the orthocenter of the triangle created by the other three.
 [/INST] Continuing with the notation and analytic-geometric approach from my comment,
$$
\eqalign{
R_1R_2 \perp R_3R_4 \iff
&
\{y=r\sin\theta\} \perp
\{x=r\cos\theta\}
\\\\&\text{True (horizontal/vertical lines)}
\\\\
R_1R_3 \perp R_2R_4 \iff
&
\overline{(a,r\sin\theta)(r\cos\theta,b)} \perp
\overline{(-a,r\sin\theta)(r\cos\theta,-b)}
\\\iff
&-1=
\frac{b-r\sin\theta}{r\cos\theta-a}\cdot\frac{-b-r\sin\theta}{r\cos\theta+a}
=\frac{r^2\sin^2\theta-b^2}{r^2\cos^2\theta-a^2}
\\&\qquad=\frac{a^2\sin^2\theta-b^2\cos^2\theta}{b^2\cos^2\theta-a^2\sin^2\theta}
\\\\&\text{True}
\\\\
R_1R_4 \perp R_2R_3 \iff
&
\overline{(a,r\sin\theta)(r\cos\theta,-b)} \perp
\overline{(-a,r\sin\theta)(r\cos\theta,b)}
\\\iff&-1=
\frac{-b-r\sin\theta}{r\cos\theta-a}\cdot\frac{b-r\sin\theta}{r\cos\theta+a}
\\\\&\text{...True?}
}
$$
Is this approach accesible for you?
Can you try the last one?
The facts we need are:
$$
\matrix{
\cos^2\theta+\sin^2\theta=1\qquad&\text{true for all }\theta
\\\\
a^2+b^2=r^2\qquad
&\matrix{\text{do you see why I set}\\\text{it up this way above?}}
\\\\
m_{PQ}=\frac{y_P-y_Q}{x_P-x_Q}
&\text{slope formula}
\\\\
-(s-t)=t-s
&\text{basic algebra}
\\\\
(s+t)(s-t)=s^2-t^2
&\text{basic algebra}
\\\\
x=r\cos\theta
&\text{is a circle with radius }r
\\
y=r\sin\theta
&\text{centered at the origin}
}
$$
</s>