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.

## Tuesday, December 16, 2008

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

Labels:
Archimedes,
book of lemmas,
circle,
diameter,
perpendicular,
secant,
tangent

Subscribe to:
Post Comments (Atom)

Let M be the center of the circle.

ReplyDeleteExtend 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)

bjhvash44@sbcglobal.net

ReplyDeleteuse 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

http://img109.imageshack.us/img109/5805/problem641.png

ReplyDeleteExtend 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

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.

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

ReplyDeleteVery nice solution !!!

Delete