Straight Edge - Only Geometric Construction

Question

Given a circle, its diameter and a given point on the diameter, find a procedure to construct a line perpendicular to the diameter using only a straight edge. The perpendicular must pass through the given point.

This question is a follow up to my previous question where the point was lying on the circle.Apparently, similar constructions cannot be made in this problem. This question discusses the case when the point lies outside the circle.

Any help will be appreciated.
Thanks.

Answer 1

The construction in the refereneced question allowes you as a side effect to reflect a point $P$ on the circle at the line $AB$ ($X\mapsto F$ with the illustration in the accepted answer).

Draw any line $\ne AB$ through $X$. Let it intersect the circle in $P$ and $Q$. Reflect $P$ at $AB$ to find $P'$. Let $AP'$ intersect $BQ$ in $C$. Then $CX\perp AB$.

Answer 2

If $UV$ and $WZ$ are two chords of $\Gamma$ through $X$, $UZ$ and $VW$ intersect on the polar line $l$ of $X$ wrt to $\Gamma$. The intersection $Y$ between $AB$ and $l$ is the inverse of $X$ with respect to $\Gamma$. If two lines through $Y$ meet $\Gamma$ at $D,E,F,G$, $DF\cap EG$ lies on the perpendicular to $AB$ through $X$.

Answer 3

If anyone wants a proof here it is:

How to contruct that? My solution is:

From $X$ draw a random straightline, and then from $B$ and $C$ draw two lines which are concurrent with $l(X)$, being $l(X)$ the line passing throw $X$, they meet at $F$, choose a point $G$ on the line $XF$, let $CG$ intersect $FB$ at $E$ and $BG$ intersect $FC$ at $H$. Now let $L$ be the point were $EH$ intersect $BC$.

At point we have found the Pole of the $l(X)$. Proof: By construction $L,B,X,C$ are harmonic conjugates. And by one of the lemmas of pole, polar and the circle is that $(L,X;B,C)=-1$.

Now the Polar can be construct from the Pole using the Brokard's Theorem. From $L$ draw a line to the circle which intersect it at $K$ and $D$. Draw the complete quadrilateral $BKDC$ and its diagonals intersect at $I$ then $IX$ is the Polar.

Observation $L$ is inverse of $X$ respect the circle and conversly