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Given a circle $C_1$, its diameter $AB$ and a Point anywhere on the plane $X$ form a method to draw perpendicular to line $AB$ passing through $X$ using only a straight edge.

I solved cases where $X$ is NOT on the circle or the line $AB$ itself with relative ease.

So my question is how to do it if $X$ lies on the circle or $AB$

How I did the first part (if $X$ not on circle or $AB$):
1. Draw $AX$ intersecting circle at $C$
2. Draw $BX$ intersecting circle at $D$
3. Draw $AD$ and $BC$ intersecting at $F$
4. $XF$ is the required perpendicular

This works because altitudes always pass through orthocenter. But this fails if $X=C,D,A,B$. How to do in those cases?

Anvit
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    Drawing a perpendicular from any other point, you get chord perpendicular to the diameter. That means that you found two points (the end-points of the chord $C,D$) and its mid point (the intersection of the chord with the diameter). Join $CX$ and intersect $AB$ at $E$. Join $ED$ and intersect the circle at $F$. Look at $XF$. –  May 11 '18 at 12:37
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    @deyore Great! And it is perpendicular because $EFX$ and $EDC$ and similar triangles and since Diameter bisects $CD$ it must bisect $XF$ too.. You wanna post that as answer? – Anvit May 11 '18 at 13:15

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