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.
 A: If you like complex numbers then the solution may be made very simple. WLOG we may assume that points $W, X, Y, Z$ all lie on the unit circle of the complex plane, and R is another point of the unit circle. We know that $WXYZ$ is a rectangle, so if, say, $W = a + i b$, then the remaining points are expressed as $X = \overline{W} = a - i b$, $Y = -W = - a - i b$, and $Z = - \overline{W} = -a + i b$.
Now let $R = p + i q$, and let us denote by $R_{A B}$ the projection of point $R$ onto line $AB$. Then by pure thought one can find that $R_{X Y} = p - i b$, $R_{Y Z} = - a + i q$, $R_{Z W} = p + i b$, and $R_{W X} = a + i q$.
We can now compute $R_{X Y} - R_{Y Z} = \overline{R} + \overline{W}$ and $R_{Z W} - R_{W X} = \overline{R} - \overline{W}$.
We need to observe that if points $A$ and $B$ lie on the unit circle, their sum $A + B$ and difference $A - B$ are orthogonal. Thus, line $(R_{X Y}, R_{Y Z})$ is perpendicular to line $(R_{Z W}, R_{W X})$
Finally, we find that line $(R_{X Y}, R_{Z W})$ is perpendicular to line $(R_{YZ}, R_{W X})$ by construction, so indeed point $R_{Z W}$ in the orthocenter of triangle $R_{XY}R_{YZ}R_{WX}$.
Similarly you may argue for the other points.
A: 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:
$$
\matrix{
\cos^2\theta+\sin^2\theta=1\qquad&\text{true for all }\theta
\\\\
a^2+b^2=r^2\qquad
&\matrix{\text{do you see why I set}\\\text{it up this way above?}}
\\\\
m_{PQ}=\frac{y_P-y_Q}{x_P-x_Q}
&\text{slope formula}
\\\\
-(s-t)=t-s
&\text{basic algebra}
\\\\
(s+t)(s-t)=s^2-t^2
&\text{basic algebra}
\\\\
x=r\cos\theta
&\text{is a circle with radius }r
\\
y=r\sin\theta
&\text{centered at the origin}
}
$$
A: Synthetic solution:
Let the projections of $R$ onto $WX, ZY, WZ$, and $XY$ be $A,B,C,D$, respectively. (In my diagram $R$ lies on the arc $XY$, for what it's worth). Let $E$ be the intersection of line $AD$ with line $CB$. 
Since trivially $CD \perp AB$ (intersecting at $R$), it's good enough to show that $ADE \perp CB$.
Claim 1: $RE \perp WY$.
Proof: Consider the Simson line of $XWY$ with respect to the point $R$. The projections onto $XW$ and $AY$ are $A$ and $D$, so the projection onto $CB$ must be $E$.
Claim 2: $WARE$ is cyclic.
Proof: $WA \perp RA$ and $RE \perp WE$.
Claim 3: $WARC$ is cyclic.
Proof: $WA \perp AR$ and $WC \perp RC$.
Claim 4: $WAEC$ is cyclic.
Proof: $WARE$ and $WARC$ are cyclic.
Claim 5: $AE \perp CE$.
Proof: $WACE$ is cyclic and $WA \perp CW$.
Claim 6: $C,E,$ and $B$ are collinear.
Proof: Simson line of $WZY$.
So $AE \perp CEB$, and $D$ is the orthocenter of $ABC$.
