Construction using a straight edge only

Question

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

I can do this if another point outside the circle is given, but my question is the degenerate case when that point lies on the circle.

Any help will be appreciated.
Thanks.

Answer 1

Here's a solution

Let $C_1 \not \in \omega$.

$H -$ orthocenter of $\triangle AC_1B$

1) $C_1H \perp AB$

2) $C_1H \cap \omega =\left\{D,E\right\}$

3) $DX \cap AB=P$

4) $PE\cap \omega = F$

5) $FX \perp AB$

Explanation for $(5)$ : Let the centre of the circle be $O$. Note that $\Delta ODP$ and $\Delta OEP$ are congruent, $\implies \angle ODP = \angle OEP$. Now, $\Delta DXO$ and $\Delta EFO$ are isosceles and congruent, so $\angle PXO = \angle PFO$ , and therefore $\Delta XPO$ and $\Delta FPO$ are congruent by SAA. If we mark the intersection point of $DE$ and $AB$ as $Z$, by the above deductions, we have that $\Delta XPZ$ and $\Delta FPZ$ are congruent, thus $\angle XZP = \angle FZP = 90^{\circ}$

The above explanation is due to Mr. Henning Makholm.

Answer 2

"how does it construct the orthocenter H using only a straightedge?"

1) $Z \in AX$;

2)$ZB \cap \omega= Y$.

Then $\angle AXB=\angle AYB =90^{\circ}$.

3) Let $AY \cap BX=H - $ orthocenter $\triangle AZB$