You have a circle <Gamma> of center O and a point A on it.
You have a line L on A and you want the other point of intersection B of
line L with circle <Gamma>.
To avoid to be stuck, follow this way and make a macro with it:
Draw through the center O the line D perpendicular to L.
Then B is the symmetric of A wrt line D.
You can also be stuck with 2 circles with a known point of intersection when
you want to draw the other.
I let you find the solutionand the macro
On 10/26/07, Jeff <cu1101@...> wrote:
> Dear Friends,
> In Hyacinthos message 11474, Jean-Pierre gave the following
> construction and proof:
> Construction of A' (cyclically for B', C')
> T = second intersection of the line BQ with the circle CPQ
> T'= second intersection of the line CQ with the circle BPQ
> then A' = CT inter BT'
> Proof : If a,b,c,p,q,a' are the complex affixes of A,B,C,P,Q,A', we
> have (p-a')/(p-c) = (p-b)/(p-q) hence the direct similitude with
> center P mapping Q to B maps C to A' and the direct similtude with
> center P mapping Q to C will map B to A'. Hence the construction
> I'm having problems making a GSP construction here. Given the four
> points B,C,P,Q I construct the circles CPQ and BPQ and the lines BQ
> and CQ respectively with no problem. But, I notice that points T and
> T' will get stuck at point Q as the point P is dragged about.
> I've seen this same situation occur in a different setting but don't
> know how to solve the problem. Can anyone help?
> Sincerely, Jeff
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