Hello, greatwhiteshark!
A fascinating problem!
There must be dozens of ways to prove it ... but I can't find <u>one</u>.
Take a segment AB and construct a semicircle on it.
Pick any point on AB and call it C. Construct semicircles on AC and CB.
Construct the perpendicular to AB at C.
The point where it intersects the big semicircle is D.
Construct segment AD and label the point where it intersects the small semicircle E.
Do the same for BD and label the point F.
Construct the line EF.
Show that line EF is the common tangent to both small semicircles.
There are so many Facts in the diagram, it should be very easy.
. . Evidently, I'm missing an important (and obvious) link.
We have right triangle ADB (inscribed in a semicircle).
Draw AE, CE, CF, and BF.
. . Then we have right triangles AEC and CFB.
All three right triangles are similar.
Let P be the center of the semicircle on AC.
Let Q be the center of the semicircle on CB.
Then EP and FQ are radii of their respective circles . . .and they are <u>parallel</u>.
We must show that EP (or FQ) is perpendicular to EF.
. . [Well, I <u>
think</u> that is our task.]
But, so far, I've had no luck . . . anyone insprired?
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
I've even noted some additional facts:
. . The median to the hypotenuse of a right triangle is one-half the hypotenuse.
. . CD is the mean proportional of AC and CB.