WLOG, suppose $XY$ is on the $x$-axis. Let us define points $X,Y$ as $(-a,0)$ and $(a,0)$, respectively. Then, $Z$ should be defined as point $(b,z)$. Additionally, point $Q=\left(q,-\frac{z}{a+b}\left(q+a\right)\right)$. For $a,b,z,q \in \Bbb R$.
We have the following equations for lines $XZ, QY$ and $ZY$
$$XZ: y=\frac{z}{a+b}\left(x+a\right)\qquad m_1=\frac{z}{a+b}\\
ZY:y=-\frac{z}{a-b}\left(x-a\right)\qquad m_2=-\frac{z}{a-b}\\
QY:y=-\frac{z(a+q)}{(a+b)(a-q)}\left(x-a\right))\qquad m_3=-\frac{z(a+q)}{(a+b)(a-q)}$$
Since we should have $\angle ZYQ=\angle ZXY$, we want the angle between lines $QY$ and $ZY$ be equal to $\angle ZXY$ Thus, we have:
$$\frac{m_2-m_3}{1+m_2m_3}=m_1$$
Substituting the slopes from the equations above, we get that $m_2$ should be:
$$m_2\to-\frac{2 a z (a+b)}{(a+b)^2 (a-q)-z^2 (a+q)}$$
However, we already know that $m_2=-\frac z{a-b}$, therefore:
$$-\frac{2 a z (a+b)}{(a+b)^2 (a-q)-z^2 (a+q)}=-\frac{z}{a-b}\\
q\to -\frac{a \left((a+b) (a-3 b)+z^2\right)}{(a+b)^2+z^2}\tag{1}$$
This gives us $q$ such that $\angle ZXY=\angle ZYQ$.
Now the line perpendicular to $XZ$ through $Q$ is defined by the line:
$$y=-\frac{a+b}{z}\left(x-q\right)+\frac{z}{a+b}\left(q+a\right)$$
Which has a zero of (i.e. $R=(x,0)$):
$$x\to \frac{z^2 (a+q)}{(a+b)^2}+q\tag{2}$$
Since we already know $q$, substituting $(1)$ in $(2)$, and we get:
$$\bbox[20px,border:1px black solid]{x\to-\frac{a (a-3 b)}{a+b}\implies
\therefore R \text{ is definitely invariant to } Z}$$
You can check this implementation.
Your basic method is sound:
- Deriving $X$ from $A$ and $B$, establish that $OX \perp XA$ (equivalently, $XB$ or $XP$).
- Recall that in a right triangle, the midpoint of its hypotenuse is equidistant from its vertices (i.e. it is its circumcentre).
- Conclude that $X$ necessarily lies on $\bigcirc M$.
- Conversely, deriving $X$ from $\bigcirc M$, extend $XP$ to a chord $AB$.
- Show that $X$ is the midpoint of this chord, so lying on $\bigcirc M$ is also sufficient.
Looking at step 4, we can say that proof 1 does this directly. Proof 2 starts at $OX \perp XP$ and shows that $X$ lies on $\bigcirc M$ as a consequence. I think this is a reason to favour proof 1, although subjectively I actually preferred proof 2!
Possible improvements are twofold: you could account for edge cases, and you could clarify your reasoning.
In the first line, allowing $O$ and $P$ to coincide is probably not worth worrying about, simply because it is trivial ($AB$ is a diameter and its midpoint is exactly $O$).
I think it is worth taking the time to consider the case where $X$ coincides with either $O$ or $P$, though, just to prove that all points on $\bigcirc M$ are valid. This means considering that $AB$ can be either a diameter of or tangent to $\bigcirc M$, so Thales’ theorem doesn’t apply.
In the second line, I stumbled at this part of your first proof:
Observe that $MX \cong MP$. For the ray from center $O$ through $X$ makes $\angle OXP$ a right angle and $\triangle OXP$ a right triangle. Then $XM$ is its median and $XM = \frac 1 2 OP = MP$.
The second and third sentences comprise a proof of the first. But they also suffice as a proof of your overall statement. So the lemma “$MX \cong MP$” is not needed.
Your second proof starts with a statement about right angles, before establishing that such an angle exists. I think it works better in this order:
A point $X$ is the midpoint of a chord $AB$ through $P$ iff $OX$ is the perpendicular bisector of $AB$. Thus any such midpoint $X$ will make $\angle OXP$ a right angle. Conversely, if $\angle OXP$ is a right angle, then $X$ is the midpoint of chord $AB$ through $P$.
By Thales' Theorem and its converse, the locus of points $X$ such that $\angle OXP$ is a right angle is precisely the circle with diameter $OP$.
(Note also the deletion of “center” from “center diameter $OP$”.)
Like your first proof, here the second and third sentences (“Thus” and “Conversely”) are a restatement of the “iff” in the first sentence. I think it would be clearer if one or the other were replaced with a justification of this relationship between a chord and its perpendicular bisector. It needn’t be a complete proof, just a pointer from this specific case to the general rule. For example:
Since $O$ and $X$ are both equidistant from $A$ and $B$, $OX$ is the perpendicular bisector of $AB$. Thus $\angle OXP$ is a right angle, and conversely, if $\angle OYP$ is a right angle, then $Y$ is the chord’s midpoint.
Best Answer
If $EA=a$ and $\angle EFA=\theta$, then $EF=EH=a\csc\theta$.
Therefore the horizontal displacement of $H$ from the line $AD$ is $a\csc \theta*\sin\theta=a$.
Therefore $H$ moves on a vertical straight line at the same distance from $AD$ as $E$ is from $A$.