Semicircles and a Tangent

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.

greatwhiteshark, are you writing a textbook or a set of contest problems?
If so, you are asking us to provide an answer key.
Gee, this is cheap labor! Shame on you!

no

This question was given by my teacher to solve over the weekend. Accroding to my math teacher, he found this question in an old textbook and decided to give us a challenge question that he said would take most of the class all weekend to solve.

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.

Click to expand...

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.

Yes...

Soroban,

Thank you for your help. OUr teacher will be giving the class weekend challenge questions from this point on. I wonder what he has instored for us next weekend.