## Tuesday, December 16, 2008

### Archimedes' Book of Lemmas, Proposition #2

Problem 641: Diameter, tangents, perpendicular
Exercise your brain. Archimedes wrote the "Book of Lemmas" more than 2200 years ago. Solve the proposition #2 (high school level) and lift up your geometry skills. 1. Let M be the center of the circle.
Extend BC and AD and let them intersect in L.
Let MC and BD intersect in K.

Then MBCD is a kite (two tangent rays, two radii), so its diagonal MC is the perpendicular bisector of its diagonal BD. So MKB=90°. We also have ADB=90° (Thales) and as orthogonals to the same line, MC and DL are parallel. Therefore BLD and BCK are similar, and we have BC/CL=BK/KD=1, or BC=CL.

Now ED and BL are parallel (orthogonal to AB), and by the intercept theorem CL/FD=AC/AF=BC/EF, and if we drop the middle part, we get EF/FD=BC/CL=1, which was claimed. (by Thomas)

2. bjhvash44@sbcglobal.net

use anom's construction to id point L
ang DAB=CDB=CBD= 1/2 arc DB=a
angALB=complement ofDAB =b
angLDC=complement of a=b
trian CLD is isosceles
CL=CD
CD=CB
CB=CL
AC is median of trian ALB It bisects all parallels to BL
BF=FE

3. http://img109.imageshack.us/img109/5805/problem641.png
Extend AD to G ( G on BC, see picture)
Triangle BDG is a right triangle ( angle ADB =90)
Triangle DCB is isosceles ( DC=CB)
Triangle CDG is isosceles ( angle CDG= angle DGC= 90- angle CBD)
So CD=CB=CG
Since DE// BG and C is midpoint of BG so F is the midpoint of DE
Peter Tran

4. If AD, BC meet at G and DE, OC meet at H then it can also be shown that BCHD is a parellogram and that BCDH is a rhombus.

5. Triangles ADE and ABD are similar, AC is symmedian of triangle ABD and consequently it is median of triangle ADE, done.

1. Very nice solution !!!