# Straight Edge - Only Geometric Construction

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.

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 $X$.

• Does this construct require that the polar line $l \perp AB$? Apr 29, 2016 at 13:49
• @Tebbe: since $X\in AB$ and $AB$ is a diameter of $\Gamma$, the polar line of $X$ is trivially perpendicular to $AB$. Apr 29, 2016 at 15:05
• @JackD'Aurizio Does your method work if $X$ is outside the circle on the extended diameter? Apr 30, 2016 at 15:30
• @Henry: it still works if $X\in AB$ but $X$ is not between $A$ and $B$. It has to be modified if $X$ is not on $AB$, but the argument through polars still gives a solution. Apr 30, 2016 at 15:32
• Anyway, if $X$ is not on $AB$, there is a simpler solution: draw $XA$ and $XB$, meeting $\Gamma$ at $C$ and $D$. $E=AD\cap BC$ is the orthocenter of $XAB$, hence $XE\perp AB$. Apr 30, 2016 at 15:38

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 $CX\perp AB$.

• It looks like this works in practice, but it deserves some kind of argument, doesn't it? Apr 28, 2016 at 17:23
• It works even if AB is not a diameter. It just requires an extra translation.
– Mick
Apr 29, 2016 at 4:30
• @HenningMakholm Of course. But the argument that $CX\perp AB$ i sthe same as for $C_1H\perp AB$ in the answer to the refernced question Apr 29, 2016 at 6:19
• @Hagen, I don't see it. It's clear to me why $C_1H$ is vertical in the previous question, but not how that argument would be modified/generalized to work here. Apr 29, 2016 at 7:33
• I can see that we could intersect $BP'$ and $AQ$ instead to meet in $C'$, and then $CC'$ would be vertical, by the orthocenter argument from the referenced question. But why does it pass through $X$? Apr 29, 2016 at 7:34

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

• Does your method work if $X$ is outside the circle, on the extended diameter? Apr 30, 2016 at 15:31