Geometry Proof Concerning Equality of Lengths on a Quadrilateral 
Quadrilateral $WXYZ$ has right angles at $\angle W$ and $\angle Y$ and an acute angle at $\angle X$. Altitudes are dropped from $X$ and $Z$ to diagonal $\overline{WY}$, meeting $\overline{WY}$ at $O$ and $P$ as shown. Prove that $OW = PY$.

The proof requires the use of similarity and special triangle parts and properties.

 A: 
$$\begin{align}
\frac{a}{p} = \frac{q}{a+b} \quad\text{and}\quad \frac{c}{q} = \frac{p}{b+c} &\quad\to\quad a(a+b) = pq = c(b+c) \\[4pt]
&\quad\to\quad a(a+b)-c(b+c) = 0 \\[4pt]
&\quad\to\quad (a-c)(a + b + c ) = 0 \\[4pt]
&\quad\to\quad a = c \quad\square
\end{align}$$
A: You can use circumcirles. Since $\angle \, XYZ = \angle \, XWZ = 90^{\circ}$ the quad $XYZW$ is inscribed in a circle $k$ and $XZ$ is a diameter whose midpoint $S$ it the center of $k$. Extend $XO$ until it intersect $k$ for the second time at point $U$. Let $\angle \, XYW = \alpha$ and $\angle \, XWY = \beta$. Then $\angle \, XZY = \angle \, WWY = \beta$ so $$\angle \, ZXY = 90^{\circ} - \beta = \angle \, WXO = \angle \, WXU$$ which yields $$YZ = WU$$ Moreover, if $\angle \, UXZ = \theta$ then 
$$\angle \, UWO = \angle \, UWY = \angle\, UXY = \beta + \theta = \angle \, ZXW = \angle \, ZYW = \angle \, ZYP$$ Consequently triangles $UWO$ and $ZYP$ are congruent and hence $OW = PY$
A: It should be clear that the said quadrilateral is circumscribed by the black circle.

Construction: 1) Extend XO to cut the circle at Q. 2) Join WQ. 3) Join ZQ. 4) Join XZ.
$\angle 1 = \angle 3$ because they both are equal to $\angle 2$. This means WQ = YZ and arc WQ = arc YZ. 
Adding the common arc ZQ to both arcs, we have the green arcs are equal. This means $\angle PYZ = \angle OWQ$. 
We now have sufficient info to conclude that $\triangle YPZ \cong \triangle WOQ$. Result follows.
