# Given Three Points...

By Alicia Rosenberger

Given three points A, B, C construct a line intersecting AC in the point Xand BC in the point Y such that AX=XY=YB

Let's start out with three points A, B', C and connect A to C and C to B', forming line segments AC and B'C respectively. We will use B' for now since the real point B is arbitrary at this point. We want to find points X on AC and Y on BC such that AX=XY=YB. so we will plot an arbitrary point x' on AC to start:

Next, we construct circles with centers A and B' such that the two circles have the same radius. So, we construct a circle with center A and radius AX' (which is arbitrary at this point). Then, we construct a circle with center B' and radius AX' (the same radius as the first circle). We label the intersection of the second circle and segment B'C, Y'. Now we can see that we have created AX'=Y'B' segments by the circle with the same radius.

Next, we want to construct a circle with center X' and radius AX' (same radius as other two circles).

If we make a line parallel to B'C intersecting AC at the point where the new circle intersects, we have achieved what we wanted, except much smaller, because we have that AX'=X'Y''=Y''B''. So what we want to do is make this bigger. We can do this by similar triangles or dilations of these ratios.

So, next I made a line through A and Y'' and marked a point where this line intersects on CB'. This is our point Y that we want. Now I can preserve the distance by that same ratio by constructing a parallel line to X'Y'' through point Y. Where this new line intersects segment AC is our point X. Now we have that AX=XY.

Now we just need to to find B so that AX=XY=YB. So, now I translated segment Y''B'' onto segment CB' at point Y.

Then I dilated segment Y''B'' by a factor of AX/AX' about center Y to get the point B.

We now have that AX=XY=YB, as desired.